∫ (1+x)/(1+x^2) dx = ?

May 2021 2 52 Report

∫ (1+x)/(1+x^2) dx = ?

Update:

So nothing fancy when showing the steps, please.

Update 3:

Thanks, I didn't know you could split the integral like that. Also, I think you made a mistake, arcsin(x) should be arctan(x).

Break it up:

= ∫ 1/(1+x²) dx + ∫ x/(1+x²) dx

= tan‾ ¹(x) + ∫ x/(1+x²) dx

For the right term use a u-substitution: u=1+x², then du=2xdx, so dx=1/(2x) du

= tan‾ ¹(x) + ½ ∫ (1/u) du

= tan‾ ¹(x) + ½ ln(x²+1)

(You don't need that absolute value round ln, since x²+1 is guaranteed to be positive)

If you split the numerator, it makes it easier...

â«1/(1 + x^2) + â«x/(1 + x^2)

arcsin(x) + â«x/(1+x^2)

u = 1 + x^2

du = 2x

1/2â«(1/u) du

1/2ln(abs(1 + x^2))

So, your answer is:

arcsin(x) + 1/2(ln(abs(1 + x^2))) + C