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

The integral of (3x+2)/(sqrt(x^2+x+2)) is

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

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

3 (x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}) + 2 ln \frac{1}{7} (4 x^{2} + 4 x + 8) + \frac{1}{7} (2 x + 1) + C

Solution steps

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

Expand

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

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

\int \frac{3 x}{x ^{2} + x + 2} d x = 3 (x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7})

\int \frac{2}{x ^{2} + x + 2} d x = 2 ln \frac{2}{7} x + \frac{1}{7} + \frac{1}{7} (4 x^{2} + 4 x + 8)

= 3 (x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}) + 2 ln \frac{2}{7} x + \frac{1}{7} + \frac{1}{7} (4 x^{2} + 4 x + 8)

Simplify

3 (x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}) + 2 ln \frac{2}{7} x + \frac{1}{7} + \frac{1}{7} (4 x^{2} + 4 x + 8) : 3 (x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}) + 2 ln \frac{1}{7} (4 x^{2} + 4 x + 8) + \frac{1}{7} (2 x + 1)

= 3 (x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}) + 2 ln \frac{1}{7} (4 x^{2} + 4 x + 8) + \frac{1}{7} (2 x + 1)

Add a constant to the solution

= 3 (x^{2} + x + 2 - \frac{1}{2} ln \frac{2}{7} x + \frac{1}{7} + \frac{4 x ^{2} + 4 x + 8}{7}) + 2 ln \frac{1}{7} (4 x^{2} + 4 x + 8) + \frac{1}{7} (2 x + 1) + C

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