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# Please help with Calculus 2 integral problem.?

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.

### 3 Answers

- 1 decade agoFavorite Answer
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!

- StevenLv 61 decade ago
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 ILv 71 decade ago
Because d/dx (x^2 + 1) = 2x, but d/dx (x^4 + 1) = 4x^3, not 3x^3. 3x^3 dx = (3/4) d(4x^3 ).

x^4 + 5 > 0 for all x, so |x^4 + 5| is unnecessary. You should just use (x^4 + 5) instead.

Source(s): I have taught for more than 40 yr, including 18 yr of AP Calc.