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

The integral of (sqrt(3x+1))/(x+2) is 2sqrt(5)(-arctan(sqrt(1/5 (3x+1)))+sqrt(1/5 (3x+1)))+C

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integral of (sqrt(3x+1))/(x+2)

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

25 (- arctan (\frac{1}{5} (3 x + 1)) + \frac{1}{5} (3 x + 1)) + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{u + 5} d u

= \int \frac{u}{u + 5} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 5} d v

= \int \frac{2 v ^{2}}{v ^{2} + 5} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{v ^{2}}{v ^{2} + 5} d v

\frac{v ^{2}}{v ^{2} + 5} = - \frac{5}{v ^{2} + 5} + 1

= 2 \cdot \int - \frac{5}{v ^{2} + 5} + 1 d v

Apply Integral Substitution

= 2 \cdot \int 5 (- \frac{1}{w ^{2} + 1} + 1) d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 25 \cdot \int - \frac{1}{w ^{2} + 1} + 1 d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 25 (- \int \frac{1}{w ^{2} + 1} d w + \int 1 d w)

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

\int 1 d w = w

= 25 (- arctan (w) + w)

Substitute back

= 25 (- arctan (\frac{3 x + 1}{5}) + \frac{3 x + 1}{5})

Simplify

25 (- arctan (\frac{3 x + 1}{5}) + \frac{3 x + 1}{5}) : 25 (- arctan (\frac{1}{5} (3 x + 1)) + \frac{1}{5} (3 x + 1))

= 25 (- arctan (\frac{1}{5} (3 x + 1)) + \frac{1}{5} (3 x + 1))

Add a constant to the solution

= 25 (- arctan (\frac{1}{5} (3 x + 1)) + \frac{1}{5} (3 x + 1)) + C

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