What is the integral of (5x-4)/(sqrt(x^2+6x+13)) ?

The integral of (5x-4)/(sqrt(x^2+6x+13)) is 5sqrt(x^2+6x+13)-19ln(1/2 |x+3+sqrt(x^2+6x+13)|)+C

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integral of (5x-4)/(sqrt(x^2+6x+13))

\int \frac{5 x - 4}{x ^{2} + 6 x + 13} d x

5 x^{2} + 6 x + 13 - 19 ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13) + C

Solution steps

\int \frac{5 x - 4}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

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

Apply u-substitution

: \int \frac{5 u - 19}{u ^{2} + 4} d u

= \int \frac{5 u - 19}{u ^{2} + 4} d u

Expand

\frac{5 u - 19}{u ^{2} + 4} : \frac{5 u}{u ^{2} + 4} - \frac{19}{u ^{2} + 4}

= \int \frac{5 u}{u ^{2} + 4} - \frac{19}{u ^{2} + 4} 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} + 4} d u - \int \frac{19}{u ^{2} + 4} d u

\int \frac{5 u}{u ^{2} + 4} d u = 5 u^{2} + 4

\int \frac{19}{u ^{2} + 4} d u = 19 ln (\frac{1}{2} u + 4 + u^{2})

= 5 u^{2} + 4 - 19 ln (\frac{1}{2} u + 4 + u^{2})

Substitute back

u = x + 3

= 5 (x + 3)^{2} + 4 - 19 ln (\frac{1}{2} x + 3 + 4 + (x + 3)^{2})

Simplify

5 (x + 3)^{2} + 4 - 19 ln (\frac{1}{2} x + 3 + 4 + (x + 3)^{2}) : 5 x^{2} + 6 x + 13 - 19 ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13)

= 5 x^{2} + 6 x + 13 - 19 ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13)

Add a constant to the solution

= 5 x^{2} + 6 x + 13 - 19 ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13) + C

\int 4 x^{3} + 3 x^{2} + 2 x + 1 + \frac{1}{x} + \frac{1}{x ^{3}} d x

x^{2} y^{^{''}} - 2 x y^{^{'}} - 4 y = 0

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

\frac{d}{d x} (\frac{2 a}{x})

What is the integral of (5x-4)/(sqrt(x^2+6x+13)) ?
The integral of (5x-4)/(sqrt(x^2+6x+13)) is 5sqrt(x^2+6x+13)-19ln(1/2 |x+3+sqrt(x^2+6x+13)|)+C