What is the integral of 2x(x^2+7)^5 ?

The integral of 2x(x^2+7)^5 is 1/6 (x^2+7)^6+C

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integral of 2x(x^2+7)^5

\int 2 x (x^{2} + 7)^{5} d x

\frac{1}{6} (x^{2} + 7)^{6} + C

Solution steps

\int 2 x (x^{2} + 7)^{5} 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} + 7)^{5} d x

Apply u-substitution

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

= 2 \cdot \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} + 7

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{6}}{6}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{6}}{6} : \frac{1}{6} (x^{2} + 7)^{6}

= \frac{1}{6} (x^{2} + 7)^{6}

Add a constant to the solution

= \frac{1}{6} (x^{2} + 7)^{6} + C

(0, 5), (3, - 5)

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

\int_{1}^{5} 10

(0, 0), (5, - 1)

\int \frac{x}{1 + x y} d x

What is the integral of 2x(x^2+7)^5 ?
The integral of 2x(x^2+7)^5 is 1/6 (x^2+7)^6+C