What is the integral of (5x-8)^7 ?

The integral of (5x-8)^7 is 1/40 (5x-8)^8+C

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integral of (5x-8)^7

\int (5 x - 8)^{7} d x

\frac{1}{40} (5 x - 8)^{8} + C

Solution steps

\int (5 x - 8)^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{5} d u

= \int \frac{u ^{7}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u^{7} d u

Apply the Power Rule

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

= \frac{1}{5} \cdot \frac{u ^{8}}{8}

Substitute back

u = 5 x - 8

= \frac{1}{5} \cdot \frac{( 5 x - 8 ) ^{8}}{8}

Simplify

\frac{1}{5} \cdot \frac{( 5 x - 8 ) ^{8}}{8} : \frac{1}{40} (5 x - 8)^{8}

= \frac{1}{40} (5 x - 8)^{8}

Add a constant to the solution

= \frac{1}{40} (5 x - 8)^{8} + C

x \to - 1 lim (\frac{1}{x - 1})

n = 1 \sum \infty 6 \cdot (0.9)^{n - 1}

\int_{1}^{1.5} \frac{1}{8} + \frac{3}{8} x d x

\frac{s}{s ^{2} + 5 s + 6}

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