What is the integral of (4x+3)/(4x^2+6x-1) ?

The integral of (4x+3)/(4x^2+6x-1) is 1/2 ln|16x^2+24x-4|+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)/(4x^2+6x-1)

\int \frac{4 x + 3}{4 x ^{2} + 6 x - 1} d x

\frac{1}{2} ln 16 x^{2} + 24 x - 4 + C

Solution steps

\int \frac{4 x + 3}{4 x ^{2} + 6 x - 1} 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 (4 x + 3)^{2} - 13

Simplify

\frac{1}{2} ln (4 x + 3)^{2} - 13 : \frac{1}{2} ln 16 x^{2} + 24 x - 4

= \frac{1}{2} ln 16 x^{2} + 24 x - 4

Add a constant to the solution

= \frac{1}{2} ln 16 x^{2} + 24 x - 4 + C

\int \frac{sin ( 2 x )}{1 + cos ( 2 x )} d x

\int csch (\frac{1}{2}) x coth (\frac{1}{2}) x d x

\int cos^{6} (8 x) sin (8 x) d x

\int \frac{1}{( x + x )} d x

\int tan^{8} (θ) sec^{2} (θ) d θ

What is the integral of (4x+3)/(4x^2+6x-1) ?
The integral of (4x+3)/(4x^2+6x-1) is 1/2 ln|16x^2+24x-4|+C