# Indefinite Integral U-Substitution

• June 8th 2010, 08:07 PM
Matt85
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.
• June 8th 2010, 08:11 PM
pickslides
$\int \frac {1+x^2} {4x} dx= \int \frac {1} {4x} +\frac{x^2}{4x}dx= \int \frac {1} {4x} +\frac{x}{4}dx$

Any better?
• June 8th 2010, 08:26 PM
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?
• June 8th 2010, 08:37 PM
harish21
Quote:

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:

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