What is the integral of 1/((x+1)sqrt(x^2+2x-8)) ?

The integral of 1/((x+1)sqrt(x^2+2x-8)) is 1/3 arcsec(1/3 (x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((x+1)sqrt(x^2+2x-8))

\int \frac{1}{( x + 1 ) x ^{2} + 2 x - 8} d x

\frac{1}{3} arcsec (\frac{1}{3} (x + 1)) + C

Solution steps

\int \frac{1}{( x + 1 ) x ^{2} + 2 x - 8} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 9} d u

= \int \frac{1}{u u ^{2} - 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (x + 1))

Simplify

= \frac{1}{3} arcsec (\frac{1}{3} (x + 1))

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{1}{3} (x + 1)) + C

\frac{x ^{3} - 1}{x ^{2}}

\int \frac{3 x - 1}{x ^{2} + 9} d x

\frac{d}{d x} (ln (g (x)))

\int \frac{x}{( x ^{2} + 9 ) ^{2}} d x

\int \frac{a}{x ^{2} + a ^{2}} d x

What is the integral of 1/((x+1)sqrt(x^2+2x-8)) ?
The integral of 1/((x+1)sqrt(x^2+2x-8)) is 1/3 arcsec(1/3 (x+1))+C