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

The integral of 1/(sqrt(x^2+2x+2)) is ln|sqrt(x^2+2x+2)+x+1|+C

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

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

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

= ln u^{2} + 1 + u

Substitute back

u = x + 1

= ln (x + 1)^{2} + 1 + x + 1

Expand

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

= ln x^{2} + 2 x + 2 + x + 1

Add a constant to the solution

= ln x^{2} + 2 x + 2 + x + 1 + C

y = (3 x - x^{2}) (3 - x - x^{2})

3 y

\frac{d}{d x} (- 3 x y ln (x y))

\int_{0}^{1} x^{10} + 1 0^{x} d x

\int 6 x e^{7 x} d x

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