What is the integral of x^7\sqrt[9]{x^8+8} ?

The integral of x^7\sqrt[9]{x^8+8} is 9/80 (x^8+8)^{10/9}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^7\sqrt[9]{x^8+8}

\int x^{7} 9 x^{8} + 8 d x

\frac{9}{80} (x^{8} + 8)^{\frac{10}{9}} + C

Solution steps

\int x^{7} 9 x^{8} + 8 d x

Apply u-substitution

: \int \frac{9 u}{8} d u

= \int \frac{9 u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int 9 u d u

Apply the Power Rule

: \frac{9}{10} u^{\frac{10}{9}}

= \frac{1}{8} \cdot \frac{9}{10} u^{\frac{10}{9}}

Substitute back

u = x^{8} + 8

= \frac{1}{8} \cdot \frac{9}{10} (x^{8} + 8)^{\frac{10}{9}}

Simplify

\frac{1}{8} \cdot \frac{9}{10} (x^{8} + 8)^{\frac{10}{9}} : \frac{9}{80} (x^{8} + 8)^{\frac{10}{9}}

= \frac{9}{80} (x^{8} + 8)^{\frac{10}{9}}

Add a constant to the solution

= \frac{9}{80} (x^{8} + 8)^{\frac{10}{9}} + C

\int \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} d x

\int cos (tan (x)) sec^{2} (x) d x

\int 68 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

\int \frac{1}{4 t ^{3}} d t

\int \frac{8}{x + 2} d x

What is the integral of x^7\sqrt[9]{x^8+8} ?
The integral of x^7\sqrt[9]{x^8+8} is 9/80 (x^8+8)^{10/9}+C