How do you integrate x3x21 using substitution?

2 Answers
Jul 29, 2016

115(3x2+2)(x21)32+C.

Explanation:

Let us take the subst. x21=t2, so, x2=t2+1, &,

2xdx=2tdt,or,xdx=tdt

Now, I=x3x21dx=x2x21xdx

=(t2+1)(t2)tdt=(t4+t2)dt=t55+t33

=t315(3t2+5)=t15t2(3t2+5)

=x2115(x21){3(x21)+5}

=115(3x2+2)(x21)32+C.

Jul 29, 2016

Here is a third solution.

Explanation:

x3x21dx=x2x21xdx

Let u=x21,

so that du=2xdx

and x2=u+1.

The integral becomes

(u+1)u1212du=12(u32+u12)du

=12[25u52+23u12]+C

=115[3u52+5u32]+C

=115u32[3u+5]+C

Back-substituting gets us

=115(x21)32[3(x21)+5]+C

=115(x21)32(3x2+2)+C