Evaluate the following integrals: x (x^2+1) x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\therefore\ 2\int\frac{\text{t dt}}{(\text{t}^2+1)\text{t}}$
$=2\Big|\frac{\text{dt}}{\text{t}^4+1}\Big|$
Dividing numerator and denominator by t²
$\text{I}=2\int\frac{\frac{\text{t}}{\text{t}^2}}{\big(\text{t}^2+\frac{1}{\text{t}^2}\big)}\text{ dt}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\big)-\Big(1-\frac{1}{\text{t}^2}\Big)}{\Big(\text{t}^2+\frac{1}{\text{t}^2}\Big)}\text{ dt}$
Let $\text{t}-\frac{1}{\text{t}}=\text{z}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dz}$ [For I^st part]
and, $\text{t}+\frac{1}{\text{t}}=\text{y}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dy}$ [For II^nd part]
$\therefore\ \text{I}=\int\frac{\text{dz}}{\text{z}^2+2}-\int\frac{\text{dy}}{\text{y}^2-2}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{z}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{y}-\sqrt{2}}{\text{y}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{2}}\tan^{1}\Big(\frac{\text{t}^2-1}{\sqrt{2}\text{t}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{x}+1-\sqrt{2\text{x}}}{\text{x}+1+\sqrt{2\text{x}}}\bigg|+\text{C}$

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