Solve \int \frac {x^3 }{(x^2+9)} dx

Question:

Solve {eq}\int \frac {x^3 }{(x^2+9)} dx {/eq}

Integration by Substitution

The integration of a function can be performed by many ways in which the method of substitution is the prominent method to integrate a function by substituting a certain group of variables with another to ease the integration.

Given data

• The given integral is: {eq}\int {\dfrac{{{x^3}}}{{{x^2} + 9}}dx} {/eq}

Rewrite the above equation,

{eq}I = \int {\dfrac{{{x^2}}}{{{x^2} + 9}}xdx} \cdots\cdots\rm{(I)} {/eq}

Let us assume that {eq}t = {x^2} + 9 {/eq}.

Differentiate the above equation.

{eq}\begin{align*} \dfrac{{dt}}{{dx}} &= 2x + 0\\ \dfrac{{dt}}{2} &= xdx \end{align*} {/eq}

From equation (I),

{eq}I = \int {\dfrac{{t - 9}}{t}\dfrac{{dt}}{2}} {/eq}

Solve the above equation.

{eq}\begin{align*} I &= \int {\dfrac{{t - 9}}{t}\dfrac{{dt}}{2}} \\ &= \dfrac{1}{2}\int {\left( {1 - \dfrac{9}{t}} \right)dt} \\ &= \dfrac{1}{2}\left( {t - 9\ln t} \right) + K \end{align*} {/eq}

Substitute the values,

{eq}I = \dfrac{1}{2}\left( {{x^2} + 9 - 9\ln \left( {{x^2} + 9} \right)} \right) + K {/eq}

Thus, the solution is {eq}\dfrac{1}{2}\left( {{x^2} + 9 - 9\ln \left( {{x^2} + 9} \right)} \right) + K {/eq}.