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

The integral of (x^2+5)^3x is 1/8 (x^2+5)^4+C

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

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

\frac{1}{8} (x^{2} + 5)^{4} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{u ^{3}}{2} d u

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{2} + 5

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 5 ) ^{4}}{4} : \frac{1}{8} (x^{2} + 5)^{4}

= \frac{1}{8} (x^{2} + 5)^{4}

Add a constant to the solution

= \frac{1}{8} (x^{2} + 5)^{4} + C

5 - 2 x

\int x (x^{2} - 1)^{99} d x

n = 0 \sum \infty \frac{1}{n + 9}

\int x^{3} + 3 d x

\frac{d y}{d x} = 2 x - y, y (0) = 1

What is the integral of (x^2+5)^3x ?
The integral of (x^2+5)^3x is 1/8 (x^2+5)^4+C