What is the integral of (x+1)/(4x^2-x-5) ?

The integral of (x+1)/(4x^2-x-5) is 1/4 ln|4x-5|+C

integral of (x+1)/(4x^2-x-5)

\int \frac{x + 1}{4 x ^{2} - x - 5} d x

\frac{1}{4} ln ∣ 4 x - 5 ∣ + C

Solution steps

\int \frac{x + 1}{4 x ^{2} - x - 5} d x

Simplify

\frac{x + 1}{4 x ^{2} - x - 5} : \frac{1}{4 x - 5}

= \int \frac{1}{4 x - 5} d x

Apply u-substitution

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 4 x - 5

= \frac{1}{4} ln ∣ 4 x - 5 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x - 5 ∣ + C

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

\int (x^{3} - 2 x + 1) e^{2 x} d x

\int (4 x^{3} + 7)^{12} x^{2} d x

\int e^{6 x} cos (6 x) d x

\int 3.17 - \frac{6.671}{x} d x

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