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

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

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

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

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

Solution steps

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

Complete the square

2 x + 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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln ∣ tan (arcsec (x + 1)) + sec (arcsec (x + 1)) ∣

Simplify

ln ∣ tan (arcsec (x + 1)) + sec (arcsec (x + 1)) ∣ : ln x^{2} + 2 x + x + 1

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

Add a constant to the solution

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

\int x^{\frac{1}{3}} + 9 d x

\int \frac{3 x ^{4} - 4 x ^{3} + 5}{x ^{2}} d x

\int (3 x^{- 2} - 4 x^{- 3}) d x

\int (3 + \frac{1}{2 x + 2}) d x

\int sec (x) sec (x) d x

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