What is the integral of (x^2+8)/(xsqrt(x^2-4)) ?

The integral of (x^2+8)/(xsqrt(x^2-4)) is sqrt(x^2-4)+4arctan(1/2 sqrt(x^2-4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2+8)/(xsqrt(x^2-4))

\int \frac{x ^{2} + 8}{x x ^{2} - 4} d x

x^{2} - 4 + 4 arctan (\frac{1}{2} x^{2} - 4) + C

Solution steps

\int \frac{x ^{2} + 8}{x x ^{2} - 4} d x

Apply u-substitution

: \int \frac{u + 8}{2 u u - 4} d u

= \int \frac{u + 8}{2 u u - 4} 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 \frac{u + 8}{u u - 4} d u

Expand

\frac{u + 8}{u u - 4} : \frac{1}{u - 4} + \frac{8}{u u - 4}

= \frac{1}{2} \cdot \int \frac{1}{u - 4} + \frac{8}{u u - 4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{1}{u - 4} d u + \int \frac{8}{u u - 4} d u)

\int \frac{1}{u - 4} d u = 2 u - 4

\int \frac{8}{u u - 4} d u = 8 arctan (\frac{1}{2} u - 4)

= \frac{1}{2} (2 u - 4 + 8 arctan (\frac{1}{2} u - 4))

Substitute back

u = x^{2}

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

Simplify

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

= x^{2} - 4 + 4 arctan (\frac{1}{2} x^{2} - 4)

Add a constant to the solution

= x^{2} - 4 + 4 arctan (\frac{1}{2} x^{2} - 4) + C

\int 5 sin^{3} (2 x) cos^{3} (2 x) d x

\int \frac{3}{5000 - 2 x} d x

\int 5 (3 - 4 x)^{\frac{2}{3}} d x

\int 1 + r^{2} r d r

\int x + 3 x + 1 d x

What is the integral of (x^2+8)/(xsqrt(x^2-4)) ?
The integral of (x^2+8)/(xsqrt(x^2-4)) is sqrt(x^2-4)+4arctan(1/2 sqrt(x^2-4))+C