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

The integral of 1/(5+4x+x^2) is arctan(x+2)+C

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

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

arctan (x + 2) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = x + 2

= arctan (x + 2)

Add a constant to the solution

= arctan (x + 2) + C

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