What is the integral of 2x(x^2+3)^7 ?

The integral of 2x(x^2+3)^7 is 1/8 (x^2+3)^8+C

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integral of 2x(x^2+3)^7

\int 2 x (x^{2} + 3)^{7} d x

\frac{1}{8} (x^{2} + 3)^{8} + C

Solution steps

\int 2 x (x^{2} + 3)^{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} + 3)^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{2} d u

= 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

: \frac{u ^{8}}{8}

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

Substitute back

u = x^{2} + 3

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 3 ) ^{8}}{8}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 3 ) ^{8}}{8} : \frac{1}{8} (x^{2} + 3)^{8}

= \frac{1}{8} (x^{2} + 3)^{8}

Add a constant to the solution

= \frac{1}{8} (x^{2} + 3)^{8} + C

x - e^{3 x}

\int \frac{3}{8 + 6 x} d x

y = \frac{1}{x}, y = \frac{1}{4} x

x \to 1 lim (\frac{x + 2}{x ^{3} + 8})

What is the integral of 2x(x^2+3)^7 ?
The integral of 2x(x^2+3)^7 is 1/8 (x^2+3)^8+C