What is the integral of 2x(x^2+6)^7 ?

The integral of 2x(x^2+6)^7 is 1/8 (x^2+6)^8+C

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integral of 2x(x^2+6)^7

\int 2 x (x^{2} + 6)^{7} d x

\frac{1}{8} (x^{2} + 6)^{8} + C

Solution steps

\int 2 x (x^{2} + 6)^{7} d x

Take the constant out:

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

= 2 \cdot \int x (x^{2} + 6)^{7} d x

Apply u-substitution

= 2 \cdot \int \frac{u ^{7}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{7} d u

Apply the Power Rule

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{8}}{8}

Substitute back

u = x^{2} + 6

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{8}}{8}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{8}}{8} : \frac{1}{8} (x^{2} + 6)^{8}

= \frac{1}{8} (x^{2} + 6)^{8}

Add a constant to the solution

= \frac{1}{8} (x^{2} + 6)^{8} + C

in v erse l a pl a ce {\frac{1}{4 s + 1}}

\int (5 sec (x) - cos (x))^{2} d x

\int (4 - x^{2}) d x

x \to 0 - lim (\frac{e ^{x}}{x ^{3}})

d er i v a t i v e \frac{y}{4}

What is the integral of 2x(x^2+6)^7 ?
The integral of 2x(x^2+6)^7 is 1/8 (x^2+6)^8+C