What is the integral of x(x^2+6)^4 ?

The integral of x(x^2+6)^4 is 1/10 (x^2+6)^5+C

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integral of x(x^2+6)^4

\int x (x^{2} + 6)^{4} d x

\frac{1}{10} (x^{2} + 6)^{5} + C

Solution steps

\int x (x^{2} + 6)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= \int \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{2} + 6

= \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{5}}{5}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{5}}{5} : \frac{1}{10} (x^{2} + 6)^{5}

= \frac{1}{10} (x^{2} + 6)^{5}

Add a constant to the solution

= \frac{1}{10} (x^{2} + 6)^{5} + C

n = 1 \sum \infty \frac{n}{e ^{3 n}}

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

\frac{d}{d x} (\frac{x ^{4}}{16} + \frac{1}{2 x ^{2}})

y = arccos (e^{3 x})

cos (3 x), 2 - cos (x)

What is the integral of x(x^2+6)^4 ?
The integral of x(x^2+6)^4 is 1/10 (x^2+6)^5+C