What is the integral of x/(sqrt(-x^2-8x-7)) ?

The integral of x/(sqrt(-x^2-8x-7)) is -sqrt(-x^2-8x-7)-4arcsin(1/3 (x+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/(sqrt(-x^2-8x-7))

\int \frac{x}{- x ^{2} - 8 x - 7} d x

- - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4)) + C

Solution steps

\int \frac{x}{- x ^{2} - 8 x - 7} d x

Complete the square

- x^{2} - 8 x - 7 : - (x + 4)^{2} + 9

= \int \frac{x}{- ( x + 4 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{u - 4}{- u ^{2} + 9} d u

= \int \frac{u - 4}{- u ^{2} + 9} d u

Expand

\frac{u - 4}{- u ^{2} + 9} : \frac{u}{- u ^{2} + 9} - \frac{4}{- u ^{2} + 9}

= \int \frac{u}{- u ^{2} + 9} - \frac{4}{- u ^{2} + 9} 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

= \int \frac{u}{- u ^{2} + 9} d u - \int \frac{4}{- u ^{2} + 9} d u

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

\int \frac{4}{- u ^{2} + 9} d u = 4 arcsin (\frac{1}{3} u)

= - - u^{2} + 9 - 4 arcsin (\frac{1}{3} u)

Substitute back

u = x + 4

= - - (x + 4)^{2} + 9 - 4 arcsin (\frac{1}{3} (x + 4))

Expand

- (x + 4)^{2} + 9 : - x^{2} - 8 x - 7

= - - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4))

Add a constant to the solution

= - - x^{2} - 8 x - 7 - 4 arcsin (\frac{1}{3} (x + 4)) + C

\int \frac{1}{2 x - 1} d x

\int x^{2} (6 x^{3} - 2)^{3} d x

\int - 9 sin (x) d x

\int \frac{13 x}{x ^{4} + 1} d x

What is the integral of x/(sqrt(-x^2-8x-7)) ?
The integral of x/(sqrt(-x^2-8x-7)) is -sqrt(-x^2-8x-7)-4arcsin(1/3 (x+4))+C