What is the integral of xsqrt(x^2-8) ?

The integral of xsqrt(x^2-8) is 1/3 (x^2-8)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xsqrt(x^2-8)

\int x x^{2} - 8 d x

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

Solution steps

\int x x^{2} - 8 d x

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = x^{2} - 8

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

Simplify

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

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

Add a constant to the solution

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

\int - 11 x d x

\int (6 x^{2} + 4 x + 2) d x

\int (x - 1) d x

\int - \frac{1}{r} d r

\int \frac{7 x + 2}{1 - 25 x ^{2}} d x

What is the integral of xsqrt(x^2-8) ?
The integral of xsqrt(x^2-8) is 1/3 (x^2-8)^{3/2}+C