Find the integral of this equation

Find the integral of x(x^2 - 1)^4 dx. :)

int: x(x^2 - 1)^4 dx

By u-substitution, we let u = x^2 - 1, and so du = 2x dx, and 1/2 du = x dx. Thus, we can re-write the integral in terms of u:

int: 1/2 u^4 du

By the power rule, this comes out to be:

1/10 u^5 + C

Plugging in what we originally called u, we get the final answer:

1/10 (x^2 - 1)^5 + C

Apply the following substitution:
u = x^2 - 1 ==> du = 2x dx.

Then, we have:
∫ x(x^2 - 1)^4 dx = 1/2 ∫ 2x(x^2 - 1)^4 dx
= 1/2 ∫ (x^2 - 1)^4 (2x dx)
= 1/2 ∫ u^4 du, by applying substitutions
= (1/2)u^(4 + 1)/(4 + 1) + C, by the Power Rule
= (1/10)u^5 + C
= (1/10)(x^2 - 1)^5 + C, since u = x^2 - 1.

I hope this helps!

∫x(x^2 - 1)^4 dx


take u = x^2 - 1

on differentiating, du/2 = x dx

now replace x by u and dx by du/2x

=1/2*∫x(u)^4 du/x

=(u^5)/10 + C

=[(x^2 - 1)^5]/10 + C

Hope u got it...

∫x(x^2 - 1)^4 dx u = x^2 - 1 du/2 = x dx 1/2*∫(u)^4 du (u^5)/10 + C (x^2 - 1)^(5)/10 + C

Use mental substitution,
integral of x(x^2 - 1)^4 dx
= integral of (1/2)(x^2 - 1)^4 d(x^2 - 4)
= (1/10)(x^2-4)^5+ c

∫x(x^2 - 1)^4 dx

u = x^2 - 1

du/2 = x dx

1/2*∫(u)^4 du

(u^5)/10 + C

(x^2 - 1)^(5)/10 + C