Practice Problems for Indefinite Integration

Integration of $ \displaystyle x^n$

$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \int{{{{x}^{n}}\ dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}+C}},n\ne -1 } \\ \hline \end{array}\end{array}$

➊       Evaluate each of the following indefinite integrals.

(a) $ \displaystyle \int{{\left( {5{{x}^{3}}-10{{x}^{2}}+4} \right)}}\ dx$

(b) $ \displaystyle \int{{\left( {{{x}^{8}}+\frac{1}{{{{x}^{8}}}}} \right)}\ }dx$

(c) $ \displaystyle \int{{\left( {3\sqrt[4]{{{{x}^{3}}}}-\frac{2}{{{{x}^{5}}}}+\frac{1}{{4\sqrt{x}}}} \right)}\ }dx$

(d) $ \displaystyle \int{{\left( {x-2} \right)(x+2)\ }}dx$

(e) $ \displaystyle \int{{2x(x+3)\ }}dx$

(f) $ \displaystyle \int{{\left( {\frac{{4{{x}^{3}}-2{{x}^{2}}+3x}}{{2x}}} \right)\ }}dx$

(g) $ \displaystyle \int{{\left( {\frac{{2x-3}}{{\sqrt[3]{x}}}} \right)\ }}dx$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \int{{(5{{x}^{3}}-10{{x}^{2}}+4)}}~dx\\\\\ \ \ \ =\ \ \displaystyle \int{{5{{x}^{3}}}}~dx-\displaystyle \int{{10{{x}^{2}}\ }}dx+\displaystyle \int{4}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{5}{4}{{x}^{4}}-\displaystyle \frac{{10}}{3}{{x}^{3}}+4x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(b)}\ \ \ \ \displaystyle \int{{\left( {{{x}^{8}}+\displaystyle \frac{1}{{{{x}^{8}}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{8}}}}~dx+\displaystyle \int{{\displaystyle \frac{1}{{{{x}^{8}}}}\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{8}}}}~dx+\displaystyle \int{{{{x}^{{-8}}}\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{9}{{x}^{9}}-\displaystyle \frac{1}{7}{{x}^{{-7}}}+C\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{x}^{9}}}}{9}-\displaystyle \frac{1}{{7{{x}^{7}}}}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(c)}\ \ \ \ \displaystyle \int{{\left( {3\sqrt[4]{{{{x}^{3}}}}-\displaystyle \frac{2}{{{{x}^{5}}}}+\displaystyle {1}{{4\sqrt{x}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{3\sqrt[4]{{{{x}^{3}}}}}}~dx-\displaystyle \int{{\displaystyle \frac{2}{{{{x}^{5}}}}\ }}dx+\displaystyle \int{{\displaystyle \frac{1}{{4\sqrt{x}}}\ }}dx\\\\\ \ \ \ =\ \ 3\displaystyle \int{{{{x}^{{\frac{3}{4}}}}}}~dx-2\displaystyle \int{{{{x}^{{-5}}}\ }}dx+\displaystyle \frac{1}{4}\displaystyle \int{{{{x}^{{-\frac{1}{2}}}}}}~dx\\\\\ \ \ \ =\ \ \displaystyle \frac{{12}}{7}{{x}^{{\frac{7}{4}}}}+\displaystyle \frac{1}{2}{{x}^{{-4}}}+\displaystyle \frac{1}{2}{{x}^{{\frac{1}{2}}}}+C\\\\\ \ \ \ =\ \ \displaystyle \frac{{12{{x}^{{\frac{7}{4}}}}}}{7}+\displaystyle \frac{1}{{2{{x}^{4}}}}+\displaystyle \frac{{\sqrt{x}}}{2}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(d)}\ \ \ \ \displaystyle \int{{\left( {x-2} \right)\left( {x+2} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {{{x}^{2}}-4} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{2}}\ }}dx-\displaystyle \int{{4\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{3}{{x}^{3}}-4x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(e)}\ \ \ \ \displaystyle \int{{2x\left( {x+3} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{2}}-6x} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{2{{x}^{2}}\ }}dx-\displaystyle \int{{6x\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{2}}\ }}dx-6\displaystyle \int{{x\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{3}{{x}^{3}}-3{{x}^{2}}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(f)}\ \ \ \ \displaystyle \int{{\left( {\displaystyle \frac{{4{{x}^{3}}-2{{x}^{2}}+3x}}{{2x}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {\displaystyle \frac{{4{{x}^{3}}}}{{2x}}-\displaystyle \frac{{2{{x}^{2}}}}{{2x}}+\displaystyle \frac{{3x}}{{2x}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{2}}-x+\displaystyle \frac{3}{2}} \right)\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{2}}\ }}dx-\displaystyle \int{{x\ }}dx+\displaystyle \frac{3}{2}\displaystyle \int{{dx}}\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{3}{{x}^{3}}-\displaystyle \frac{1}{2}{{x}^{2}}+\displaystyle \frac{3}{2}x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(g)}\ \ \ \ \displaystyle \int{{\left( {\displaystyle \frac{{2x-3}}{{\sqrt[3]{x}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{{ \frac{2}{3}}}}-3{{x}^{{-\frac{1}{3}}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{{\frac{2}{3}}}}\ }}dx-3\displaystyle \int{{{{x}^{{-\frac{1}{3}}}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{6}{5}{{x}^{{\frac{5}{3}}}}-\displaystyle \frac{9}{2}{{x}^{{\frac{2}{3}}}}+C\end{array}$

➋       Given that $ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}-4x-5$ and $ \displaystyle y=-4$ when $ \displaystyle x=-2,$ find the value of $ \displaystyle y$ when $ \displaystyle x=1.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}-4x-5\\\\\therefore \ \ \ \ dy=\left( {3{{x}^{2}}-4x-5} \right)dx\\\\\therefore \ \ \ \ y=\displaystyle \int{{\left( {3{{x}^{2}}-4x-5} \right)}}dx\\\\\ \ \ \ \ \ y={{x}^{3}}-2{{x}^{2}}-5x+C\\\\\ \ \ \ \ \ \text{When}\ x=-2,\ y=-4\\\\\therefore \ \ \ \ {{\left( {-2} \right)}^{3}}-2{{\left( {-2} \right)}^{2}}-5\left( {-2} \right)+C=-4\\\\\therefore \ \ \ \ C=2\\\\\therefore \ \ \ \ y={{x}^{3}}-2{{x}^{2}}-5x+2\\\\\therefore \ \ \ \ \text{When}\ x=1,\ y=-4\end{array}$

➌       If $ \displaystyle {g}'(x)= 3x^2-2x-3$ and $ \displaystyle g(-1)=4,$ factorize g(x) completely.

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$ \displaystyle \begin{array}{l}\ \ \ \ {g}'(x)=3{{x}^{2}}-2x-3\\\\\therefore \ \ g(x)=\displaystyle \int{{{g}'(x)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \int{{\left( {3{{x}^{2}}-2x-3} \right)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ = {{x}^{3}}-{{x}^{2}}-3x+C\\\\\ \ \ \ \text{Since}\ g(-1)=4,\\\\\ \ \ \ -1-1+3+C=4\\\\\therefore \ \ C=3\\\\\therefore \ \ g(x)={{x}^{3}}-{{x}^{2}}-3x+3\\\\\ \ \ \ g(1)=1-1-3+3=0\end{array}$ $ \displaystyle \therefore \ \ x-1$ is a factor of $ \displaystyle g(x).$ $\displaystyle \begin{matrix}\begin{array}{r} \left. 1 \right) \\ ~ \end{array} & \underline { \begin{array}{rrrr} 1&-1&-3&3\\ ~~~~&1&0&-3 \end{array} } \\ ~ & \begin{array}{rrrr} \ \ { 1 }& \ \ \ { 0 }& { -3 }&\ \ \color{red}{ 0 } \end{array}\end{matrix}$ $ \displaystyle \begin{array}{l}\therefore \ \ g(x)=(x-1)({{x}^{2}}-3)\\\\\therefore \ \ g(x)=(x-1)(x-\sqrt{3})(x+\sqrt{3})\end{array}$

➍       Given that the curve which has gradient $ \displaystyle \frac{dy}{dx}=2x^2 + 7x$ at any point on the curve passes through the origin, determine the equation of the curve.

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$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{{dy}}{{dx}}=2{{x}^{2}}+7x\\\\\therefore \ \ dy=\left( {2{{x}^{2}}+7x} \right)dx\\\\\therefore \ \ y=\displaystyle \int{{\left( {2{{x}^{2}}+7x} \right)}}\ dx\\\\\therefore \ \ y=\displaystyle \frac{2}{3}{{x}^{3}}+\displaystyle \frac{7}{2}{{x}^{2}}+C\end{array}$ Since the curve passes through the origin, when $ \displaystyle x=0$ and $ \displaystyle y=0$. $ \displaystyle \begin{array}{l}\therefore \ \ 0=\displaystyle \frac{2}{3}\left( 0 \right)+\displaystyle \frac{7}{2}\left( 0 \right)+C\\\\\therefore \ \ C=0\\\\\therefore \ \ y=\displaystyle \frac{2}{3}{{x}^{3}}+\displaystyle \frac{7}{2}{{x}^{2}}\end{array}$

➎       The rate of change of $ \displaystyle A$ with respect to $ \displaystyle r$ is given by $ \displaystyle \frac{{dA}}{{dr}}=6r+1$. If $ \displaystyle A = 3$ when $ \displaystyle r = 1$, find $ \displaystyle A$ in terms of $ \displaystyle r$.

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{dA}}{{dr}}=6r+1\\\\\therefore \ dA=\left( {6r+1} \right)dr\\\\\therefore \ A=\displaystyle \int{{\left( {6r+1} \right)}}\ dr\\\\\therefore \ A=3{{r}^{2}}+r+C\end{array}$ When $ \displaystyle r=1$ and $ \displaystyle A=3$. $ \displaystyle \begin{array}{l}\therefore \ 3=3{{\left( 1 \right)}^{2}}+\left( 1 \right)+C\\\\\therefore \ C=-1\\\\\therefore \ A=3{{r}^{2}}+r-1\end{array}$

➏       If the curve which has gradient $ \displaystyle \frac{dy}{dx}=kx-1$ at any point on the curve cuts the $ \displaystyle x$-axis at $ \displaystyle x=-3$ and $ \displaystyle x=4$, determine the value of $ \displaystyle k$ and hence find the equation of the curve.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=kx-1\\\\\therefore \ \ \ dy=\left( {kx-1} \right)dx\\\\\therefore \ \ \ y=\displaystyle \int{{\left( {kx-1} \right)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{k}{2}{{x}^{2}}-x+C\end{array}$ Since the curve cuts the $ \displaystyle x$-axis at $ \displaystyle x=-3$ and $ \displaystyle x=4$ $ \displaystyle \begin{array}{l}\ \ \ \ \ \displaystyle \frac{k}{2}{{\left( {-3} \right)}^{2}}-\left( {-3} \right)+C=0\\\\\therefore \ \ \ 9k+6+2C=0\ ---(1)\\\\\ \ \ \ \ \displaystyle \frac{k}{2}{{\left( 4 \right)}^{2}}-\left( 4 \right)+C=0\\\\\therefore \ \ \ 16k-8+2C=0\ ---(2)\\\\\ \ \ \ \ \text{By}\ (2)-(1),\\\\\ \ \ \ \ 7k-14=0\\\\\therefore \ \ \ k=2\\\\\therefore \ \ \ 9(2)+6+2C=0\ \\\\\therefore \ \ \ C=-12\\\\\therefore \ \ \ y={{x}^{2}}-x-12\ \ \end{array}$

➐       If $ \displaystyle f''\left( x \right) = 15\sqrt x + 5{x^3} + 6$, $ \displaystyle f\left( 1 \right) = - \frac{5}{4}$ and $ \displaystyle f\left( 4 \right) = 404$, find the expression for $ \displaystyle f(x)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {f}''\left( x \right)=15\sqrt{x}+5{{x}^{3}}+6\\\\\therefore \ \ {f}'\left( x \right)=\displaystyle \int{{{f}''\left( x \right)}}\ dx\\\\\therefore \ \ {f}'\left( x \right)=\displaystyle \int{{\left( {15\sqrt{x}+5{{x}^{3}}+6} \right)}}\ dx\\\\\therefore \ \ {f}'\left( x \right)=10{{x}^{{\displaystyle \frac{3}{2}}}}+\displaystyle \frac{5}{4}{{x}^{4}}+6x+c\\\\\therefore \ \ f\left( x \right)=\displaystyle \int{{{f}'\left( x \right)}}\ dx\\\\\therefore \ \ f\left( x \right)=\displaystyle \int{{\left( {10{{x}^{{ \frac{3}{2}}}}+\displaystyle \frac{5}{4}{{x}^{4}}+6x+c} \right)}}\ dx\\\\\therefore \ \ f\left( x \right)=4{{x}^{{ \frac{5}{2}}}}+\displaystyle \frac{1}{4}{{x}^{5}}+3{{x}^{2}}+cx+d\\\\\ \ \ \ f(1)=-\displaystyle \frac{5}{4}\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 4+\displaystyle \frac{1}{4}+3+c+d=-\displaystyle \frac{5}{4}\\\\\therefore \ \ c+d=-\displaystyle \frac{{17}}{2}\ \ ---(1)\\\\\ \ \ \ f(4)=404\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 128+231+48+4c+d=404\\\\\therefore \ \ 4c+d=-28\ \ ---(2)\\\\\therefore \ \ c=-\displaystyle \frac{{13}}{2}\ \operatorname{and}\ d=-2\\\\\therefore \ \ f\left( x \right)=4{{x}^{{\frac{5}{2}}}}+\displaystyle \frac{1}{4}{{x}^{5}}+3{{x}^{2}}-\displaystyle \frac{{13}}{2}x-2\end{array}$

➑       Given that a ball moves along a grove with the velocity $ \displaystyle v=\frac{{ds}}{{dt}}=32t-2$ ft/s and at the time $ \displaystyle t=\frac{1}{2}\ \text{s}$ the ball is $ \displaystyle 4\ \text{ft}$ away from the starting point. Express the position of the ball as a function of time.

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$ \displaystyle \begin{array}{l}\ \ \ \ v=\displaystyle \frac{{ds}}{{dt}}=32t-2\\\\\therefore \ \ ds=\left( {32t-2} \right)dt\\\\\therefore \ \ s=\displaystyle \int{{\left( {32t-2} \right)}}\ dt\\\\\therefore \ \ s=16{{t}^{2}}-2t+C\\\\\ \ \ \ \text{When}\ t=\displaystyle \frac{1}{2},\ s=4\\\\\therefore \ \ 16{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-2\left( {\displaystyle \frac{1}{2}} \right)+C=4\\\\\therefore \ \ C=1\\\\\therefore \ \ s=16{{t}^{2}}-2t+1\end{array}$

Integration of $ \displaystyle (ax + b)^n$

$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \int{{{{{\left( {ax+b} \right)}}^{n}}}}\ dx=\displaystyle \frac{{{{{\left( {ax+b} \right)}}^{{n+1}}}}}{{a\left( {n+1} \right)}}+C,} \\ {\text{where}\ n\ne -1,\ a\ne 0 \ \ \ \ \ \ \ \ \ }\\ \hline \end{array}\end{array}$

➒       Integrate each of the following with respect to $ \displaystyle x$.

(a) $ \displaystyle (2x+1)^6$

(b) $ \displaystyle \frac{2}{{\sqrt{{3x-2}}}}$

(c) $ \displaystyle (5x-11)^{10}$

(d) $ \displaystyle {{\left( {\frac{3}{{4x-1}}} \right)}^{3}}$

(e) $ \displaystyle \frac{2}{{5{{{\left( {3x-1} \right)}}^{4}}}}$

(f) $ \displaystyle \sqrt{{{{{\left( {5x+2} \right)}}^{3}}}}$

(g) $ \displaystyle \frac{5}{{2\sqrt{{{{{\left( {3x-1} \right)}}^{7}}}}}}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \ \displaystyle \int{{{{{(2x+1)}}^{6}}}}\ dx\ \\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{7\times 2}}{{(2x+1)}^{7}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{14}}{{(2x+1)}^{7}}+C\\\\\\\\\text{(b)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{\sqrt{{3x-2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ 2\displaystyle \int{{{{{\left( {3x-2} \right)}}^{{- \frac{1}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ 2\times \displaystyle \frac{{{{{\left( {3x-2} \right)}}^{{\frac{1}{2}}}}}}{{\displaystyle \frac{1}{2}\times 3}}\ +C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{4}{3}\sqrt{{3x-2}}+C\\\\\\\\\text{(c)}\ \ \ \ \ \displaystyle \int{{{{{(5x-11)}}^{{10}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{11\times 5}}{{(5x-11)}^{{10}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{55}}{{(5x-11)}^{{10}}}+C\\\\\\\\\text{(d)}\ \ \ \ \ \displaystyle \int{{{{{\left( {\displaystyle \frac{3}{{4x-1}}} \right)}}^{3}}}}\ dx\\\\\ \ \ \ \ =\ \ 27\displaystyle \int{{{{{\left( {4x-1} \right)}}^{{-3}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{27}}{{-2\times 4}}{{\left( {4x-1} \right)}^{{-2}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{{27}}{{8{{{\left( {4x-1} \right)}}^{2}}}}+C\\\\\\\\\text{(e)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{5{{{\left( {3x-1} \right)}}^{4}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{5}\displaystyle \int{{{{{\left( {3x-1} \right)}}^{{-4}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{{5(-3)(3)}}{{\left( {3x-1} \right)}^{{-3}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{2}{{45{{{\left( {3x-1} \right)}}^{3}}}}+C\\\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \int{{\sqrt{{{{{\left( {5x+2} \right)}}^{3}}}}}}\ dx\\\\\ \ \ \ \ =\ \displaystyle \int{{{{{\left( {5x+1} \right)}}^{{\frac{3}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{\displaystyle \frac{5}{2}\times 5}}{{\left( {5x+1} \right)}^{{\displaystyle \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{{25}}{{\left( {5x+1} \right)}^{{\frac{5}{2}}}}+C\\\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{5}{{2\sqrt{{{{{\left( {3x-1} \right)}}^{7}}}}}}}}\ dx\\\\\ \ \ \ \ =\ \displaystyle \frac{5}{2}\displaystyle \int{{{{{\left( {3x-1} \right)}}^{{-\frac{7}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{5}{2}{{\left( {3x-1} \right)}^{{- \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{5}{{2\left( {-\displaystyle \frac{5}{2}} \right)3}}{{\left( {3x-1} \right)}^{{- \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{1}{{3{{{\left( {3x-1} \right)}}^{{ \frac{5}{2}}}}}}+C\end{array}$

➓       Find the equation of the curve which cuts the $ \displaystyle y$-axis at $ \displaystyle (0, 4)$ and for which $ \displaystyle \frac{{dy}}{{dx}}={{(2x+1)}^{3}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{dy}}{{dx}}={{(2x+1)}^{3}}\\\\\therefore \ \ dy={{(2x+1)}^{3}}dx\\\\\therefore \ \ y=\displaystyle \int{{{{{(2x+1)}}^{3}}}}\ dx\\\\\therefore \ \ y=\displaystyle \frac{1}{{4\times 2}}{{(2x+1)}^{4}}+C\\\\\therefore \ \ y=\displaystyle \frac{1}{8}{{(2x+1)}^{4}}+C\end{array}$ Since the curve passes through $ \displaystyle (0, 4)$, $ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{1}{8}{{(2(0)-1)}^{4}}+C=4\\\\\therefore \ \ \ C=\displaystyle \frac{{31}}{8}\\\\\therefore \ \ y=\displaystyle \frac{1}{8}{{(2x+1)}^{4}}+\displaystyle \frac{{31}}{8}\\\\\therefore \ \ y=\displaystyle \frac{1}{8}\left[ {{{{(2x+1)}}^{4}}+31} \right]\end{array}$