What is the integral of (4x-7)/(sqrt(x^2+6x+5)) ?

The integral of (4x-7)/(sqrt(x^2+6x+5)) is 4sqrt(x^2+6x+5)-19ln(1/2 |sqrt(x^2+6x+5)+x+3|)+C

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integral of (4x-7)/(sqrt(x^2+6x+5))

\int \frac{4 x - 7}{x ^{2} + 6 x + 5} d x

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

Solution steps

\int \frac{4 x - 7}{x ^{2} + 6 x + 5} d x

Complete the square

x^{2} + 6 x + 5 : (x + 3)^{2} - 4

= \int \frac{4 x - 7}{( x + 3 ) ^{2} - 4} d x

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 3

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (4x-7)/(sqrt(x^2+6x+5)) ?
The integral of (4x-7)/(sqrt(x^2+6x+5)) is 4sqrt(x^2+6x+5)-19ln(1/2 |sqrt(x^2+6x+5)+x+3|)+C