$\int \frac{1}{x^{2} + a^{2}} d x$ $\int1/(x^2+a^2)dx$ ?

Question

"Let $a$ $a$ be a constant, and evaluate the integral."

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Answer 1 · 2018-04-13T07:26:24

See process below

$\int \frac{1}{x^{2} + a^{2}} d x$ $int1/(x^2+a^2)dx$

Lets do the change $x = a tan θ$ $x=atantheta$ . With this change we have

$d x = a {sec}^{2} θ d θ$ $dx=asec^2theta d theta$ . Put in the integral

$int(cancelasec^2theta d theta)/(a^cancel2(tan^2theta+1))$

But $tan^2theta+1=sec^2theta$ . The we have

$int(cancelacancelsec^2theta d theta)/(a^cancel2cancelsec^2theta)=1/aint d theta=1/atheta +C$

Undoing the change $theta= arctan(x/a)$ . We have finally

$int1/(x^2+a^2)dx=1/a arctan(x/a)+C$