What is the integral of (4x+3)/(2x^2+3x-12) ?

The integral of (4x+3)/(2x^2+3x-12) is ln|16x^2+24x-96|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x+3)/(2x^2+3x-12)

\int \frac{4 x + 3}{2 x ^{2} + 3 x - 12} d x

ln 16 x^{2} + 24 x - 96 + C

Solution steps

\int \frac{4 x + 3}{2 x ^{2} + 3 x - 12} d x

Apply u-substitution

: \int \frac{2 u}{u ^{2} - 105} d u

= \int \frac{2 u}{u ^{2} - 105} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{u ^{2} - 105} d u

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

= 2 \cdot \frac{1}{2} ln (4 x + 3)^{2} - 105

Simplify

2 \cdot \frac{1}{2} ln (4 x + 3)^{2} - 105 : ln 16 x^{2} + 24 x - 96

= ln 16 x^{2} + 24 x - 96

Add a constant to the solution

= ln 16 x^{2} + 24 x - 96 + C

\int (x - \frac{1}{x ^{2}}) d x

\int \frac{x - x + 5}{x ^{4}} d x

\int 4 x 5 x - 2 d x

\int 4 e^{x} sec (e^{x} + 1) d x

\int cos^{4} (a x) d x

What is the integral of (4x+3)/(2x^2+3x-12) ?
The integral of (4x+3)/(2x^2+3x-12) is ln|16x^2+24x-96|+C