Parciace

Jak počítat integrály ve tvaru In=∫1(x2+1)n dx\displaystyle I_n = \int \frac{1}{{\left(x^2 + 1\right)}^n} \,{\mathrm dx}

I1=∫1x2+1 dx=xx2+1+2∫x2(x2+1)2 dx=xx2+1+2∫x2(x2+1)2 dx=xx2+1+2∫x2+1(x2+1)2 dx−2∫1(x2+1)2 dx=xx2+1+2I1−2I2\begin{align*}I_1 = \int \frac{1}{x^2 + 1} \,{\mathrm dx} &= \frac{x}{x^2 + 1} + 2 \int \frac{x^2}{{\left(x^2 + 1\right)}^2} \,{\mathrm dx} \\&= \frac{x}{x^2 + 1} + 2 \int \frac{x^2}{{\left(x^2 + 1\right)}^2} \,{\mathrm dx} \\&= \frac{x}{x^2 + 1} + 2 \int \frac{x^2 + 1}{{\left(x^2 + 1\right)}^2} \,{\mathrm dx} - 2 \int \frac{1}{{\left(x^2 + 1\right)}^2} \,{\mathrm dx} \\&= \frac{x}{x^2 + 1} + 2 I_1 - 2 I_2\end{align*} I2=12(xx2+1+arctan⁡(x))+CI_2 = \frac{1}{2} {\left(\frac{x}{x^2 + 1} + \arctan{\left(x\right)}\right)} + C