What is the integral of (2x+3)/((x^2+3x-8)^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+3)/((x^2+3x-8)^2)

\int \frac{2 x + 3}{( x ^{2} + 3 x - 8 ) ^{2}} d x

- \frac{1}{x ^{2} + 3 x - 8} + C

Solution steps

\int \frac{2 x + 3}{( x ^{2} + 3 x - 8 ) ^{2}} d x

Apply u-substitution

: \int \frac{8 u}{( u ^{2} - 41 ) ^{2}} d u

= \int \frac{8 u}{( u ^{2} - 41 ) ^{2}} d u

Take the constant out:

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

= 8 \cdot \int \frac{u}{( u ^{2} - 41 ) ^{2}} d u

Apply u-substitution

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

= 8 \cdot \int \frac{1}{2 v ^{2}} d v

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 8 \cdot \frac{1}{2} (- \frac{1}{v})

Substitute back

= 8 \cdot \frac{1}{2} (- \frac{1}{( 2 x + 3 ) ^{2} - 41})

Simplify

8 \cdot \frac{1}{2} (- \frac{1}{( 2 x + 3 ) ^{2} - 41}) : - \frac{1}{x ^{2} + 3 x - 8}

= - \frac{1}{x ^{2} + 3 x - 8}

Add a constant to the solution

= - \frac{1}{x ^{2} + 3 x - 8} + C

\int 2 y sin (y^{2}) d y

\int x e^{- (a x + b)^{2}} d x

\int x ln (2 x) - x d x

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

\int x^{4} y cos (y^{2}) d y

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