Help calc!!?

Question

find indefinite integral
1. integral x/ (x^2 +1) dx
2. integral (lnx)^2/x dx


thanks thanks

Answer 1

# 1
∫ [ x /(x^2 + 1)] dx

Let (x^2 + 1) = z, then 2xdx = dz
So the given integral
= 0.5∫ [dz /z]
= 0.5 ln(z) + c
= 0.5ln(x^2 + 1) + c, where c = arbitrary constant of integration


# 2
∫ [(lnx)^2/x] dx

Let ln(x) = z, then dx/x = dz
So the given integral becomes
∫ [(z)^2 dz]
= (z^3)/3 + c
= [{ln(x)}^3]/3 + c, where c = arbitrary constant of integration

Answer 2

Let me try.
Derivative of x^2 + 1 = 2x
So, multiply and divide be 2.
1/2 * integral 2x dx / (x^2 + 1)
integral = 1/2 ln ( x^2 + 1) + 1/2 C

Next one, I'll have to figure.

Answer 3

For the first one use "u-subsititution"...
let u=x^2 + 1
so du is 2x
so you have 1/2 [integral] 1/u du
then find the antiderivative i don't quit remember but i think...
if f(x)=1/u then F(x)= ln u so you have....

1/2 [ln(x^2 + 1)] +C

check that work but I think its right but its been awhile since i had calculus so i don't know for sure about that antiderivative.

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integral (ln x)^2 /x dx

bring the 2 down in front so
2 integral (ln x)/x
let u=ln x
so du= 1/x
so...
2 integral u du
so....
2 * 1/2 u^2
so....
u^2 + C
so....
(ln x)^2 + C

Answer 4

I am horrible at integrals, but I know that this is for sure a U substitution problem. Good luck!