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

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

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 2 x - 1

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

Add a constant to the solution

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

\frac{d}{d x} (e^{- x} cos (y))

\frac{d}{d x} ((1 + x^{3})^{5})

\frac{\partial}{\partial y} (e^{y} sin (x) y)

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

\frac{d y}{d t} = - \frac{6}{( 2000 + 6 t )}

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