What is the integral of (5x+2)/(sqrt(x^2+2x+10)) ?

The integral of (5x+2)/(sqrt(x^2+2x+10)) is 5sqrt(x^2+2x+10)-3ln(1/3 |x+1+sqrt(x^2+2x+10)|)+C

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integral of (5x+2)/(sqrt(x^2+2x+10))

\int \frac{5 x + 2}{x ^{2} + 2 x + 10} d x

5 x^{2} + 2 x + 10 - 3 ln (\frac{1}{3} x + 1 + x^{2} + 2 x + 10) + C

Solution steps

\int \frac{5 x + 2}{x ^{2} + 2 x + 10} d x

Complete the square

x^{2} + 2 x + 10 : (x + 1)^{2} + 9

= \int \frac{5 x + 2}{( x + 1 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{5 u - 3}{u ^{2} + 9} d u

= \int \frac{5 u - 3}{u ^{2} + 9} d u

Expand

\frac{5 u - 3}{u ^{2} + 9} : \frac{5 u}{u ^{2} + 9} - \frac{3}{u ^{2} + 9}

= \int \frac{5 u}{u ^{2} + 9} - \frac{3}{u ^{2} + 9} 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{5 u}{u ^{2} + 9} d u - \int \frac{3}{u ^{2} + 9} d u

\int \frac{5 u}{u ^{2} + 9} d u = 5 u^{2} + 9

\int \frac{3}{u ^{2} + 9} d u = 3 ln (\frac{1}{3} u + 9 + u^{2})

= 5 u^{2} + 9 - 3 ln (\frac{1}{3} u + 9 + u^{2})

Substitute back

u = x + 1

= 5 (x + 1)^{2} + 9 - 3 ln (\frac{1}{3} x + 1 + 9 + (x + 1)^{2})

Simplify

5 (x + 1)^{2} + 9 - 3 ln (\frac{1}{3} x + 1 + 9 + (x + 1)^{2}) : 5 x^{2} + 2 x + 10 - 3 ln (\frac{1}{3} x + 1 + x^{2} + 2 x + 10)

= 5 x^{2} + 2 x + 10 - 3 ln (\frac{1}{3} x + 1 + x^{2} + 2 x + 10)

Add a constant to the solution

= 5 x^{2} + 2 x + 10 - 3 ln (\frac{1}{3} x + 1 + x^{2} + 2 x + 10) + C

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What is the integral of (5x+2)/(sqrt(x^2+2x+10)) ?
The integral of (5x+2)/(sqrt(x^2+2x+10)) is 5sqrt(x^2+2x+10)-3ln(1/3 |x+1+sqrt(x^2+2x+10)|)+C