What is the integral of x^2(2x^3+4)^5 ?

The integral of x^2(2x^3+4)^5 is 1/36 (2x^3+4)^6+C

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integral of x^2(2x^3+4)^5

\int x^{2} (2 x^{3} + 4)^{5} d x

\frac{1}{36} (2 x^{3} + 4)^{6} + C

Solution steps

\int x^{2} (2 x^{3} + 4)^{5} d x

Apply u-substitution

: \int \frac{16 u ( u ^{\frac{3}{2}} + 2 ) ^{5}}{x} d u

= \int \frac{( 2 u + 4 ) ^{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 (2 u + 4)^{5} d u

Apply u-substitution

: \int \frac{v ^{5}}{2} d v

= \frac{1}{3} \cdot \int \frac{v ^{5}}{2} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{v ^{6}}{6}

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{( 2 x ^{3} + 4 ) ^{6}}{6}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{( 2 x ^{3} + 4 ) ^{6}}{6} : \frac{1}{36} (2 x^{3} + 4)^{6}

= \frac{1}{36} (2 x^{3} + 4)^{6}

Add a constant to the solution

= \frac{1}{36} (2 x^{3} + 4)^{6} + C

\int \frac{x}{k} d x

x^{2} + x^{3} 1

(4 t^{2} - t) (t^{3} - 8 t^{2} + 12)

y = sin (x), y = 2 sin^{2} (x), x = 0, x = \frac{π}{2}

x \to π - lim (\frac{x}{cos ( x ) + 1})

What is the integral of x^2(2x^3+4)^5 ?
The integral of x^2(2x^3+4)^5 is 1/36 (2x^3+4)^6+C