How do you use Integration by Substitution to find #intx/(x^2+1)dx#?

2 Answers
Mar 16, 2018

Answer:

#intx/(x^2+1)dx=1/2ln(x^2+1)+"c"#

Explanation:

We want of find #intx/(x^2+1)dx#.

We make the natural substitution #u=x^2+1# and #du=2xdx#.

So

#intx/(x^2+1)dx=1/2int1/udu=1/2lnabsu= 1/2lnabs(x^2+1)+"c"#

We notice that #x^2+1>0# so we don't need to take the absolute value. So the final answer is

#1/2ln(x^2+1)+"c"#

Mar 19, 2018

Answer:

You don't. You simply write down the answer.

Explanation:

#int x/(x^2+1)\ dx#
#=(1/2)int (2x)/(x^2+1)#
The integral is of the form #int (f'(x))/f(x)\ dx=ln|f(x)|#
So you can just write down the answer, dropping the #||#, as #x^2+1>0#:

#1/2ln(x^2+1)+c#
or alternatively #lnsqrt(x^2+1)+c#
With practice you can do this type in one step at sight!