What is the integral of 4x(2x^2+1)^7 ?

The integral of 4x(2x^2+1)^7 is 1/8 (2x^2+1)^8+C

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integral of 4x(2x^2+1)^7

\int 4 x (2 x^{2} + 1)^{7} d x

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

Solution steps

\int 4 x (2 x^{2} + 1)^{7} d x

Take the constant out:

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

= 4 \cdot \int x (2 x^{2} + 1)^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{4} d u

= 4 \cdot \int \frac{u ^{7}}{4} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int u^{7} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{4} \cdot \frac{u ^{8}}{8}

Substitute back

u = 2 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int 6 x^{4} - 5 x^{3} + 5 x^{2} + C d x

\frac{d y}{d x} = e^{2 y - x}

y = (7 x - 4) (2 + 6 x)

x \to \infty lim (arctan (1 - x))

x \to 0 + lim (arctan (ln (x)))

What is the integral of 4x(2x^2+1)^7 ?
The integral of 4x(2x^2+1)^7 is 1/8 (2x^2+1)^8+C