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

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

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

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

- \frac{1}{2 ( x ^{2} + 4 x - 1 )} + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{( u ^{2} - 5 ) ^{2}} d u

= \int \frac{u}{( u ^{2} - 5 ) ^{2}} d u

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

= \frac{1}{2} (- \frac{1}{( x + 2 ) ^{2} - 5})

Simplify

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

= - \frac{1}{2 ( x ^{2} + 4 x - 1 )}

Add a constant to the solution

= - \frac{1}{2 ( x ^{2} + 4 x - 1 )} + C

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