# Indefinite Integral U-Substitution

• Jun 8th 2010, 07:07 PM
Matt85
Indefinite Integral U-Substitution
$\displaystyle \int \frac {1+x^2} {4x} dx$

I'm having alot of trouble with this problem. I've tried using both $\displaystyle 1+x^2$ and $\displaystyle 4x$ as U but I can't get the x's to cancel.
• Jun 8th 2010, 07:11 PM
pickslides
$\displaystyle \int \frac {1+x^2} {4x} dx= \int \frac {1} {4x} +\frac{x^2}{4x}dx= \int \frac {1} {4x} +\frac{x}{4}dx$

Any better?
• Jun 8th 2010, 07:26 PM
Matt85
So,

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

Yeah?
• Jun 8th 2010, 07:37 PM
harish21
Quote:

Originally Posted by Matt85
So,

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

Yeah?

the answer you got is correct but:

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