What is the integral of ((x+1))/(sqrt(3x^2+6x+5)) ?

The integral of ((x+1))/(sqrt(3x^2+6x+5)) is 1/3 sqrt(3x^2+6x+5)+C

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integral of ((x+1))/(sqrt(3x^2+6x+5))

\int \frac{( x + 1 )}{3 x ^{2} + 6 x + 5} d x

\frac{1}{3} 3 x^{2} + 6 x + 5 + C

Solution steps

\int \frac{x + 1}{3 x ^{2} + 6 x + 5} d x

Expand

\frac{x + 1}{3 x ^{2} + 6 x + 5} : \frac{x}{3 x ^{2} + 6 x + 5} + \frac{1}{3 x ^{2} + 6 x + 5}

= \int \frac{x}{3 x ^{2} + 6 x + 5} + \frac{1}{3 x ^{2} + 6 x + 5} 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{x}{3 x ^{2} + 6 x + 5} d x + \int \frac{1}{3 x ^{2} + 6 x + 5} d x

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

\int \frac{1}{3 x ^{2} + 6 x + 5} d x = \frac{1}{3} ln \frac{3}{2} (x + 1) + \frac{3 x ^{2} + 6 x + 5}{2}

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

Add similar elements:

- \frac{1}{3} ln \frac{3 x ^{2} + 6 x + 5}{2} + \frac{3}{2} (x + 1) + \frac{1}{3} ln \frac{3 x ^{2} + 6 x + 5}{2} + \frac{3}{2} (x + 1) = 0

= \frac{1}{3} 3 x^{2} + 6 x + 5

Add a constant to the solution

= \frac{1}{3} 3 x^{2} + 6 x + 5 + C

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What is the integral of ((x+1))/(sqrt(3x^2+6x+5)) ?
The integral of ((x+1))/(sqrt(3x^2+6x+5)) is 1/3 sqrt(3x^2+6x+5)+C