What is the integral of x^2(x^3+1)^{3/4} ?

The integral of x^2(x^3+1)^{3/4} is 4/21 (x^3+1)^{7/4}+C

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integral of x^2(x^3+1)^{3/4}

\int x^{2} (x^{3} + 1)^{\frac{3}{4}} d x

\frac{4}{21} (x^{3} + 1)^{\frac{7}{4}} + C

Solution steps

\int x^{2} (x^{3} + 1)^{\frac{3}{4}} d x

Apply u-substitution

= \int \frac{u ^{\frac{3}{4}}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int u^{\frac{3}{4}} d u

Apply the Power Rule

= \frac{1}{3} \cdot \frac{4}{7} u^{\frac{7}{4}}

Substitute back

u = x^{3} + 1

= \frac{1}{3} \cdot \frac{4}{7} (x^{3} + 1)^{\frac{7}{4}}

Simplify

\frac{1}{3} \cdot \frac{4}{7} (x^{3} + 1)^{\frac{7}{4}} : \frac{4}{21} (x^{3} + 1)^{\frac{7}{4}}

= \frac{4}{21} (x^{3} + 1)^{\frac{7}{4}}

Add a constant to the solution

= \frac{4}{21} (x^{3} + 1)^{\frac{7}{4}} + C

What is the integral of x^2(x^3+1)^{3/4} ?
The integral of x^2(x^3+1)^{3/4} is 4/21 (x^3+1)^{7/4}+C