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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} - \frac{1}{u ^{2} - 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\int 9.8

\int \frac{1 + ln ( x )}{( x ln ( x ) )} d x

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