What is the integral of x^2sqrt((x^3+1)) ?

The integral of x^2sqrt((x^3+1)) is 2/9 (x^3+1)^{3/2}+C

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integral of x^2sqrt((x^3+1))

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

\frac{2}{9} (x^{3} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x^{3} + 1 d x

Apply u-substitution

: \int \frac{u}{3} d u

= \int \frac{u}{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 d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{3} + 1

= \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (x^{3} + 1)^{\frac{3}{2}} : \frac{2}{9} (x^{3} + 1)^{\frac{3}{2}}

= \frac{2}{9} (x^{3} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x^{3} + 1)^{\frac{3}{2}} + C

x \to 0 lim (sin (x) + cos (x))

y = \frac{x ^{3} + 8}{x}

\int \frac{16}{16 + e ^{x}} d x

f (u) = \frac{1}{u}

n = 1 \sum \infty \frac{( ( - 1 ) ^{n} )}{n}

What is the integral of x^2sqrt((x^3+1)) ?
The integral of x^2sqrt((x^3+1)) is 2/9 (x^3+1)^{3/2}+C