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

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

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

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

arctan (x + 1) + C

Solution steps

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

Complete the square

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

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

= arctan (x + 1)

Add a constant to the solution

= arctan (x + 1) + C

\int (4 x^{5} - x^{3} - 5 x) d x

\int - 8 x^{7} y^{- 5} + 6 x^{6} y^{- 4} d y

\int (ln (x))^{10} \frac{1}{x} d x

y^{^{'''}} - y^{^{''}} + 8 y^{^{'}} - 8 y = 0

25 x e^{- x}

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