What is the integral of (4-7x)/(sqrt(x^2+2x-8)) ?

The integral of (4-7x)/(sqrt(x^2+2x-8)) is 11ln(1/3 |x+sqrt(x^2+2x-8)+1|)-7sqrt(x^2+2x-8)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4-7x)/(sqrt(x^2+2x-8))

\int \frac{4 - 7 x}{x ^{2} + 2 x - 8} d x

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

Solution steps

\int \frac{4 - 7 x}{x ^{2} + 2 x - 8} d x

Expand

\frac{4 - 7 x}{x ^{2} + 2 x - 8} : \frac{4}{x ^{2} + 2 x - 8} - \frac{7 x}{x ^{2} + 2 x - 8}

= \int \frac{4}{x ^{2} + 2 x - 8} - \frac{7 x}{x ^{2} + 2 x - 8} d x

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

\int \frac{4}{x ^{2} + 2 x - 8} d x = 4 ln (\frac{1}{3} x^{2} + 2 x - 8 + x + 1)

\int \frac{7 x}{x ^{2} + 2 x - 8} d x = 7 (x^{2} + 2 x - 8 - ln (\frac{1}{3} x^{2} + 2 x - 8 + x + 1))

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

Simplify

4 ln (\frac{1}{3} x^{2} + 2 x - 8 + x + 1) - 7 (x^{2} + 2 x - 8 - ln (\frac{1}{3} x^{2} + 2 x - 8 + x + 1)) : 11 ln (\frac{1}{3} x + x^{2} + 2 x - 8 + 1) - 7 x^{2} + 2 x - 8

= 11 ln (\frac{1}{3} x + x^{2} + 2 x - 8 + 1) - 7 x^{2} + 2 x - 8

Add a constant to the solution

= 11 ln (\frac{1}{3} x + x^{2} + 2 x - 8 + 1) - 7 x^{2} + 2 x - 8 + C

\int 9 4 x^{10} d x

\int x (5 x^{2} - 3)^{7} d x

\int e^{4 x} - x^{- 1} d x

\int e^{x} sin (x) cos (x) d x

\int \frac{9}{( t - 8 ) ^{2}} d t

What is the integral of (4-7x)/(sqrt(x^2+2x-8)) ?
The integral of (4-7x)/(sqrt(x^2+2x-8)) is 11ln(1/3 |x+sqrt(x^2+2x-8)+1|)-7sqrt(x^2+2x-8)+C