What is the integral of (x+2)/(x^2+3x-4) ?

The integral of (x+2)/(x^2+3x-4) is 2(1/4 ln|4x^2+12x-16|-1/20 (ln| 8/5+2/5 x|-ln|-2/5+2/5 x|))+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)/(x^2+3x-4)

\int \frac{x + 2}{x ^{2} + 3 x - 4} d x

2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

Solution steps

\int \frac{x + 2}{x ^{2} + 3 x - 4} d x

Complete the square

x^{2} + 3 x - 4 : (x + \frac{3}{2})^{2} - \frac{25}{4}

= \int \frac{x + 2}{( x + \frac{3}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

= \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

Take the constant out:

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

= 2 \cdot \int \frac{2 u + 1}{4 u ^{2} - 25} d u

Expand

\frac{2 u + 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} + \frac{1}{4 u ^{2} - 25}

Apply the Sum Rule:

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

= 2 (\int \frac{2 u}{4 u ^{2} - 25} d u + \int \frac{1}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1))

Substitute back

u = x + \frac{3}{2}

= 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1))

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - \frac{1}{20} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 - \frac{1}{20} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

\frac{\partial}{\partial x} (\frac{4}{36 - x ^{2}})

(2) (- 3.1)

\int cos^{7} (8 - x) sin (8 - x) d x

y^{^{'}} - 4 y = 9 e^{7 t}

\int tan^{2} (8 x) sec^{4} (8 x) d x

What is the integral of (x+2)/(x^2+3x-4) ?
The integral of (x+2)/(x^2+3x-4) is 2(1/4 ln|4x^2+12x-16|-1/20 (ln| 8/5+2/5 x|-ln|-2/5+2/5 x|))+C