# Math Help - Indefinite Integral U-Substitution

1. ## Indefinite Integral U-Substitution

$\int \frac {1+x^2} {4x} dx$

I'm having alot of trouble with this problem. I've tried using both $1+x^2$ and $4x$ as U but I can't get the x's to cancel.

2. $\int \frac {1+x^2} {4x} dx= \int \frac {1} {4x} +\frac{x^2}{4x}dx= \int \frac {1} {4x} +\frac{x}{4}dx$

Any better?

3. So,

$\int \frac {1} {4x} + \frac {x} {4} = \int \frac {1} {4} (x)^{-1} + \frac {1} {4} x= \frac {1} {4} ln |x| + \frac {1}{8}x^2 + c$

Yeah?

4. Originally Posted by Matt85
So,

$\int \frac {1} {4x} + \frac {x} {4} = \int \frac {1} {4} (x)^{-1} + \frac {1} {4} x= \frac {1} {4} ln |x| + \frac {1}{8}x^2 + c$

Yeah?
the answer you got is correct but:

$\int \frac {1} {4x} + \frac {x} {4} = \frac{1}{4} ln(x) + \frac{1}{4} \times \frac{x^{(1+1)}}{(1+1)}+C= \frac{1}{4} ln(x) + \frac{x^2}{8} + C$