What is the integral of (1+x^3)^5x^2 ?

The integral of (1+x^3)^5x^2 is 1/18 (1+x^3)^6+C

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integral of (1+x^3)^5x^2

\int (1 + x^{3})^{5} x^{2} d x

\frac{1}{18} (1 + x^{3})^{6} + C

Solution steps

\int (1 + x^{3})^{5} x^{2} d x

Apply u-substitution

: \int \frac{u ^{5}}{3} d u

= \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = 1 + x^{3}

= \frac{1}{3} \cdot \frac{( 1 + x ^{3} ) ^{6}}{6}

Simplify

\frac{1}{3} \cdot \frac{( 1 + x ^{3} ) ^{6}}{6} : \frac{1}{18} (1 + x^{3})^{6}

= \frac{1}{18} (1 + x^{3})^{6}

Add a constant to the solution

= \frac{1}{18} (1 + x^{3})^{6} + C

\frac{d}{d y} (x y)

x^{2} y^{^{'}} + 4 x y = \frac{sin ( 3 x )}{x ^{2}}

\int \frac{2 x - 1}{x - 1} d x

\int \frac{1}{t ^{2} 9 - t ^{2}} d t

\frac{d ^{2}}{d y ^{2}} (cos (x y^{2}))

What is the integral of (1+x^3)^5x^2 ?
The integral of (1+x^3)^5x^2 is 1/18 (1+x^3)^6+C