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

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

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

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

- \frac{1}{2 ( x ^{2} + 2 x - 1 )} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

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

= \frac{1}{2} (- \frac{1}{x ^{2} + 2 x - 1})

Simplify

\frac{1}{2} (- \frac{1}{x ^{2} + 2 x - 1}) : - \frac{1}{2 ( x ^{2} + 2 x - 1 )}

= - \frac{1}{2 ( x ^{2} + 2 x - 1 )}

Add a constant to the solution

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

x \to 2 - lim (5 x^{5} - 9 x^{4} + 5)

cos (2 t - 2) e^{3 t}

\int_{0}^{1.75} 1500 x d x

\int \frac{1}{x ^{3} - 4 x ^{2} + 8 x} d x

x \to 3 lim (\frac{1}{6 - 2 x})

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