How do you integrate by substitution $\int x^{2} {(x^{3} - 1)}^{4} d x$ $int x^2(x^3-1)^4 dx$ ?

Question

How do you integrate by substitution $\int x^{2} {(x^{3} - 1)}^{4} d x$ $int x^2(x^3-1)^4 dx$ ?

1 Answer

Impact of this question

7837 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-05T07:36:50

THe answer is $\frac{{(x^{3} - 1)}^{5}}{15} + C$ $(x^3-1)^5/15+C$

Explanation:

Let $u = (x^{3} - 1)$ $u=(x^3-1)$
Then $d u = 3 x^{2} d x$ $du=3x^2dx$ $\Rightarrow$ $=>$ $\frac{d u}{3} = x^{2} d x$ $(du)/3=x^2dx$
So $\int x^{2} {(x^{3} - 1)}^{4} d x = \int \frac{u^{4} d u}{3}$ $intx^2(x^3-1)^4dx=int(u^4du)/3$
$= \frac{1}{3} \cdot \frac{u^{5}}{5} = \frac{u^{5}}{15}$ $=1/3*u^5/5=u^5/15$
$:.intx^2(x^3-1)^4dx=(x^3-1)^5/15+C$

Science

Math

Humanities

... and beyond

How do you integrate by substitution $\int x^{2} {(x^{3} - 1)}^{4} d x$ $int x^2(x^3-1)^4 dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate by substitution int x^2(x^3-1)^4 dx∫x2(x3−1)4dxint x^2(x^3-1)^4 dx?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate by substitution $\int x^{2} {(x^{3} - 1)}^{4} d x$ $int x^2(x^3-1)^4 dx$ ?