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

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

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

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

- \frac{1}{x + 1} + C

Solution steps

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

Complete the square

1 + 2 x + x^{2} : (x + 1)^{2}

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 1

= - \frac{1}{x + 1}

Add a constant to the solution

= - \frac{1}{x + 1} + C

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