Mathematics

$$\int {\dfrac{{{x^5}}}{{{x^2}\, + \,9}}} \,dx$$


SOLUTION

Consider the given integral.

$$I=\int{\dfrac{{{x}^{5}}}{{{x}^{2}}+9}}dx$$


Let,

$$ t={{x}^{2}}+9 $$

$$ dt=2xdx $$

Therefore,

$$ I=\dfrac{1}{2}\int{\dfrac{{{\left( t-9 \right)}^{2}}}{t}}dt $$

$$ I=\dfrac{1}{2}\int{\dfrac{{{t}^{2}}+81-18t}{t}}dt $$

$$ I=\dfrac{1}{2}\int{\left( t+\dfrac{81}{t}-18 \right)}dt $$

$$ I=\dfrac{1}{2}\left[ \dfrac{{{t}^{2}}}{2}+81\ln \left( t \right)-18t \right]+C $$

 

Put the value of $$t$$, we get

$$I=\dfrac{1}{2}\left[ \dfrac{{{\left( {{x}^{2}}+9 \right)}^{2}}}{2}+81\ln \left( {{x}^{2}}+9 \right)-18\left( {{x}^{2}}+9 \right) \right]+C$$

 

Hence, the value of this expression is $$\dfrac{1}{2}\left[ \dfrac{{{\left( {{x}^{2}}+9 \right)}^{2}}}{2}+81\ln \left( {{x}^{2}}+9 \right)-18\left( {{x}^{2}}+9 \right) \right]+C$$

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