What is the integral of (4x+3)/(sqrt(x^2+4x+8)) ?

The integral of (4x+3)/(sqrt(x^2+4x+8)) is 4sqrt(x^2+4x+8)-5ln(1/2 |x+2+sqrt(x^2+4x+8)|)+C

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integral of (4x+3)/(sqrt(x^2+4x+8))

\int \frac{4 x + 3}{x ^{2} + 4 x + 8} d x

4 x^{2} + 4 x + 8 - 5 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8) + C

Solution steps

\int \frac{4 x + 3}{x ^{2} + 4 x + 8} d x

Expand

\frac{4 x + 3}{x ^{2} + 4 x + 8} : \frac{4 x}{x ^{2} + 4 x + 8} + \frac{3}{x ^{2} + 4 x + 8}

= \int \frac{4 x}{x ^{2} + 4 x + 8} + \frac{3}{x ^{2} + 4 x + 8} d x

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 x}{x ^{2} + 4 x + 8} d x + \int \frac{3}{x ^{2} + 4 x + 8} d x

\int \frac{4 x}{x ^{2} + 4 x + 8} d x = 4 (x^{2} + 4 x + 8 - 2 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8))

\int \frac{3}{x ^{2} + 4 x + 8} d x = 3 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8)

= 4 (x^{2} + 4 x + 8 - 2 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8)) + 3 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8)

Simplify

4 (x^{2} + 4 x + 8 - 2 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8)) + 3 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8) : 4 x^{2} + 4 x + 8 - 5 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8)

= 4 x^{2} + 4 x + 8 - 5 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8)

Add a constant to the solution

= 4 x^{2} + 4 x + 8 - 5 ln (\frac{1}{2} x + 2 + x^{2} + 4 x + 8) + C

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What is the integral of (4x+3)/(sqrt(x^2+4x+8)) ?
The integral of (4x+3)/(sqrt(x^2+4x+8)) is 4sqrt(x^2+4x+8)-5ln(1/2 |x+2+sqrt(x^2+4x+8)|)+C