Question

Calculate the following integral \[\int\limits \tan^{-1} x dx \]

Answer 1

integration by parts
u = tan^(-1)x
du / dx = 1/(1+x^2)

dv = 1
v = x

Answer 2

\[= xtan^{-1}x - \int x * \frac{1}{1+x^2}dx\]

Answer 3

\[ = xtan^{-1}x - \int\frac{x}{1+x^2}dx\]

Answer 4

for \[\int \frac{x}{1+x^2}\]
use u-sub
u = 1+ x^2
du/dx = 2x

Answer 5

0.5ln1+x^2

Answer 6

I've got the answer it says so thank you!