Question #d53c9

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Answer 1 · 2017-04-07T16:38:59

#1/4 x^4 - 1/2x^2 + 1/2 ln |x^2+1| + C#

Using long division: #(x^5)/(x^2+1) = x^3-x + x/(x^2+1)#

So #int (x^5)/(x^2+1) dx = int (x^3-x) dx + int x/(x^2+1) dx#

For the last piece, let #u = x^2+1, du = 2x dx#

#int (x^5)/(x^2+1) dx = 1/4 x^4 - 1/2x^2 + 1/2 int (du)/u#

#int (x^5)/(x^2+1) dx = 1/4 x^4 - 1/2x^2 + 1/2 ln |x^2+1| + C#

Simplified:

#int (x^5)/(x^2+1) dx = 1/4 x^2(x^2-2)+ 1/2 ln |x^2+1| + C#

Answer 2 · 2017-04-07T16:47:13

#intx^5/(x^2+1)dx = x^4/4 -x^2/2 + 1/2ln(x^2+1)+C#

Given:

#intx^5/(x^2+1)dx = #

Let #u = x^2", then "du = 2xdx#:

#1/2intu^2/(u+1)du=#

#1/2int(u^2+u-u)/(u+1)du=#

#1/2int(u^2+u)/(u+1)-(u)/(u+1)du=#

#1/2intu -(u)/(u+1)du=#

#1/2intu -(u+ 1 - 1)/(u+1)du=#

#1/2intu -(u+ 1)/(u+1) + 1/(u+1)du=#

#1/2intu -1 + 1/(u+1)du=#

#1/2(u^2/2 -u + ln|u+1|)+ C=#

Reverse the substitution:

#1/2(x^4/2 -x^2 + ln(x^2+1))+ C=#

#x^4/4 -x^2/2 + 1/2ln(x^2+1)+C#