What is the integral of x/(sqrt(x^2+2x+25)) ?

The integral of x/(sqrt(x^2+2x+25)) is sqrt(x^2+2x+25)-ln(1/(2sqrt(6))|x+1+sqrt(x^2+2x+25)|)+C

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integral of x/(sqrt(x^2+2x+25))

\int \frac{x}{x ^{2} + 2 x + 25} d x

x^{2} + 2 x + 25 - ln (\frac{1}{2 6} x + 1 + x^{2} + 2 x + 25) + C

Solution steps

\int \frac{x}{x ^{2} + 2 x + 25} d x

Complete the square

x^{2} + 2 x + 25 : (x + 1)^{2} + 24

= \int \frac{x}{( x + 1 ) ^{2} + 24} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{2} + 24} d u

= \int \frac{u - 1}{u ^{2} + 24} d u

Expand

\frac{u - 1}{u ^{2} + 24} : \frac{u}{u ^{2} + 24} - \frac{1}{u ^{2} + 24}

= \int \frac{u}{u ^{2} + 24} - \frac{1}{u ^{2} + 24} 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} + 24} d u - \int \frac{1}{u ^{2} + 24} d u

\int \frac{u}{u ^{2} + 24} d u = u^{2} + 24

\int \frac{1}{u ^{2} + 24} d u = ln (\frac{1}{2 6} u + 24 + u^{2})

= u^{2} + 24 - ln (\frac{1}{2 6} u + 24 + u^{2})

Substitute back

u = x + 1

= (x + 1)^{2} + 24 - ln (\frac{1}{2 6} x + 1 + 24 + (x + 1)^{2})

Simplify

(x + 1)^{2} + 24 - ln (\frac{1}{2 6} x + 1 + 24 + (x + 1)^{2}) : x^{2} + 2 x + 25 - ln (\frac{1}{2 6} x + 1 + x^{2} + 2 x + 25)

= x^{2} + 2 x + 25 - ln (\frac{1}{2 6} x + 1 + x^{2} + 2 x + 25)

Add a constant to the solution

= x^{2} + 2 x + 25 - ln (\frac{1}{2 6} x + 1 + x^{2} + 2 x + 25) + C

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What is the integral of x/(sqrt(x^2+2x+25)) ?
The integral of x/(sqrt(x^2+2x+25)) is sqrt(x^2+2x+25)-ln(1/(2sqrt(6))|x+1+sqrt(x^2+2x+25)|)+C