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

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

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

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

- ln 3 x + 1 + 1 + ln 3 x + 1 - 1 + 2 3 x + 1 + C

Solution steps

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

Apply u-substitution

= \int \frac{2 u ^{2}}{u ^{2} - 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

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

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

= 2 (- \int \frac{1}{- u ^{2} + 1} - 1 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

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

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

= 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u))

Substitute back

u = 3 x + 1

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

Simplify

2 (- (\frac{ln 3 x + 1 + 1}{2} - \frac{ln 3 x + 1 - 1}{2} - 3 x + 1)) : - ln 3 x + 1 + 1 + ln 3 x + 1 - 1 + 2 3 x + 1

= - ln 3 x + 1 + 1 + ln 3 x + 1 - 1 + 2 3 x + 1

Add a constant to the solution

= - ln 3 x + 1 + 1 + ln 3 x + 1 - 1 + 2 3 x + 1 + C

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