What is the integral of (x^2+8)^7x ?

The integral of (x^2+8)^7x is 1/16 (x^2+8)^8+C

Popular >

integral of (x^2+8)^7x

\int (x^{2} + 8)^{7} x d x

\frac{1}{16} (x^{2} + 8)^{8} + C

Solution steps

\int (x^{2} + 8)^{7} x d x

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 8

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 8 ) ^{8}}{8} : \frac{1}{16} (x^{2} + 8)^{8}

= \frac{1}{16} (x^{2} + 8)^{8}

Add a constant to the solution

= \frac{1}{16} (x^{2} + 8)^{8} + C

\int \frac{arctan ( \frac{1}{x} )}{x ^{4}} d x

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

\int \frac{x ^{4}}{1 - 4 x ^{10}} d x

\int x (5 x^{2} - 1)^{3} d x

\int \frac{2 x ^{2} + x - 1}{1 + x ^{2}} d x

What is the integral of (x^2+8)^7x ?
The integral of (x^2+8)^7x is 1/16 (x^2+8)^8+C