What is the integral of (x+2)/(x^2+4x-9) ?

The integral of (x+2)/(x^2+4x-9) is 1/2 ln|x^2+4x-9|+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+4x-9)

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

\frac{1}{2} ln x^{2} + 4 x - 9 + C

Solution steps

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

Complete the square

x^{2} + 4 x - 9 : (x + 2)^{2} - 13

= \int \frac{x + 2}{( x + 2 ) ^{2} - 13} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 13} d u

= \int \frac{u}{u ^{2} - 13} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln (x + 2)^{2} - 13

Simplify

\frac{1}{2} ln (x + 2)^{2} - 13 : \frac{1}{2} ln x^{2} + 4 x - 9

= \frac{1}{2} ln x^{2} + 4 x - 9

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x - 9 + C

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

\int \frac{9 x ^{2} - 49}{x ^{4}} d x

\int 4 x^{2} - 1 d x

lo g_{10} (x)

\int \frac{1}{2 u ^{3}} d u

What is the integral of (x+2)/(x^2+4x-9) ?
The integral of (x+2)/(x^2+4x-9) is 1/2 ln|x^2+4x-9|+C