1. ## Integration

Find ∫x tan inverse (x^2) dx, for x from 0 to 1.

2. $\displaystyle\int_0^1 x\arctan x^2dx=\int_0^1\left(\frac{x^2}{2}\right)'\arctan x^2dx=$

$\displaystyle=\frac{x^2}{2}\arctan x^2|_0^1 -\int_0^1\frac{x^3}{1+x^4}dx=$

$\displaystyle\frac{\pi}{8}-\frac{1}{3}\ln(1+x^4)\left|_0^1\right.=\frac{\pi}{ 8}-\frac{1}{3}\ln 2$

3. Hello! Thank you for helping, may I know how did you get from

$\displaystyle\frac{x^2}{2}\arctan x^2|_0^1 -\int_0^1\frac{x^3}{1+x^4}dx$ to $\displaystyle\frac{\pi}{8}-\frac{1}{3}\ln(1+x^4)\left|_0^1\right.$ ??

Thank you!

4. $\frac{x^{2}}{2}\arctan x^{2}\bigg|_{0}^{1}=\frac{1^{2}}{2}\arctan 1-\frac{0^{2}}{2}\arctan 0=\frac{\pi }{8}.$

Those are things you should know since you're evaluating definite integrals.

As for the last integral, it's an $\frac14$ instead an $\frac13.$

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Just for fun: $\int_{0}^{1}{x\arctan x^{2}\,dx}=\int_{0}^{1}{\int_{0}^{x^{2}}{\frac{x}{ 1+y^{2}}\,dy}\,dx}=\int_{0}^{1}{\int_{\sqrt{y}}^{1 }{\frac{x}{1+y^{2}}\,dx}\,dy},$ hence $\int_{0}^{1}{x\arctan x^{2}\,dx=\frac{1}{2}\int_{0}^{1}{\frac{1-y}{1+y^{2}}\,dy}=}\frac{\pi }{8}-\frac{\ln 2}{4}.$