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

The integral of 1/(sqrt(x^2+4x+5)) is ln|sqrt(x^2+4x+5)+x+2|+C

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

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

ln x^{2} + 4 x + 5 + x + 2 + C

Solution steps

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

Complete the square

x^{2} + 4 x + 5 : (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 = ln u^{2} + 1 + u

= ln u^{2} + 1 + u

Substitute back

u = x + 2

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

Expand

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

= ln x^{2} + 4 x + 5 + x + 2

Add a constant to the solution

= ln x^{2} + 4 x + 5 + x + 2 + C

\int \frac{x ^{5} + x ^{2} + 2}{x ^{3} - x} d x

\int e^{- 3 x^{- 5}} x^{- 6} d x

\int sin^{3} (x) (cos^{2} (x)) d x

e^{- x}, - 0.75

f^{^{''}} (x) = x

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