Why ∫(2x) / (x^2 + 1)dx

= ln l x^2 + 1 l + c

But, ∫(3x^3) / (x^4 + 5)dx

= 3/4 ln l x^4 + 5 l + c and not just ln l x^4 + 5 l + c?

How do you know when to use just ln l l + c and when to use number ln l l + c? Thank you.

Relevance

use sub t=x^2 +1

dt = 2x.dx

so integral is,

∫1/t.dt

= ln(t) + C

and t = x^2 +1

so

= ln(x^2 +1) +C

Now you do the other one!

• Steven
Lv 6

It has to do with the substitution you're using.

for ∫(2x) / (x^2 + 1)dx we let u = x^2 + 1. this gives du = 2x dx --> or dx = 1/(2x) du

substitute u for x^2 + 1 and 1/(2x) du for dx, (simplify) and you get

∫1/u du which is ln |u| + c

back-substitute gives ln |x^2 + 1| + c

now, follow the same process for ∫(3x^3) / (x^4 + 5)dx

let u = x^4 + 5. then du = 4x^3 dx --> or dx = 1/(4x^3) du

substitute u for x^4 + 5 and 1/(4x^3) du for dx, (simplify) and you get

∫(3/4)*1/u du which is (3/4)*∫1/u du which is (3/4)*ln |u| + c

after back-substituting you get 3/4*ln |x^4 + 5| + c

hope that helps

• Ed I
Lv 7