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

The integral of x(5x^2+4)^3 is 1/40 (5x^2+4)^4+C

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

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

\frac{1}{40} (5 x^{2} + 4)^{4} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{3}}{10} d u

= \int \frac{u ^{3}}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{10} \cdot \frac{u ^{4}}{4}

Substitute back

u = 5 x^{2} + 4

= \frac{1}{10} \cdot \frac{( 5 x ^{2} + 4 ) ^{4}}{4}

Simplify

\frac{1}{10} \cdot \frac{( 5 x ^{2} + 4 ) ^{4}}{4} : \frac{1}{40} (5 x^{2} + 4)^{4}

= \frac{1}{40} (5 x^{2} + 4)^{4}

Add a constant to the solution

= \frac{1}{40} (5 x^{2} + 4)^{4} + C

(1, 2.5), (4, 1)

\int 3 x^{3} - 12 x d x

\frac{4}{s ^{2} ( s + 2 )}

y^{^{''}} - 4 y = 16 x e^{2 x}

\frac{d y}{d t} + tan (t) y = 3 sin (t), y (0) = 1

What is the integral of x(5x^2+4)^3 ?
The integral of x(5x^2+4)^3 is 1/40 (5x^2+4)^4+C