How do you evaluate the definite integral int (x^3+2x)/(x^2+1)x3+2xx2+1 from [0, 2][0,2]?

1 Answer
May 7, 2018

The answer is =2.8047=2.8047

Explanation:

First, calculate the indefinite integral by substitution

Let u=x^2+1u=x2+1, =>, du=du=2xdx#

Therefore,

The indefinite integral is

I=int((x^3+2x)dx)/(x^2+1)=int((x^2+2)xdx)/(x^2+1)I=(x3+2x)dxx2+1=(x2+2)xdxx2+1

=1/2int((u+1)du)/u=12(u+1)duu

=1/2int(1+1/u)du=12(1+1u)du

=1/2(u+lnu)=12(u+lnu)

=1/2(x^2+1)+1/2ln(x^2+1)+C=12(x2+1)+12ln(x2+1)+C

The definite integral is

int_0^2((x^3+2x)dx)/(x^2+1)=[1/2(x^2+1)+1/2ln(x^2+1)]_0^220(x3+2x)dxx2+1=[12(x2+1)+12ln(x2+1)]20

=(5/2+1/2ln(5))-(1/2+0)

=2+1/2ln(5)

=2.8047