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

The integral of 1/(sqrt(x(3x+5))) is 1/(sqrt(3))ln((|sqrt(36x^2+60x)+6x+5|)/5)+C

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integral of 1/(sqrt(x(3x+5)))

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

\frac{1}{3} ln (\frac{36 x ^{2} + 60 x + 6 x + 5}{5}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u ( u - 5 )} d u

= \int \frac{1}{3 u ( u - 5 )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ( u - 5 )} d u

Complete the square

u (u - 5) : (u - \frac{5}{2})^{2} - \frac{25}{4}

= \frac{1}{3} \cdot \int \frac{1}{( u - \frac{5}{2} ) ^{2} - \frac{25}{4}} d u

Apply u-substitution

: \int \frac{2}{4 v ^{2} - 25} d v

= \frac{1}{3} \cdot \int \frac{2}{4 v ^{2} - 25} d v

Take the constant out:

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

= \frac{1}{3} \cdot 2 \cdot \int \frac{1}{4 v ^{2} - 25} d v

Apply Trigonometric Substitution

: \int \frac{sec ( w )}{2} d w

= \frac{1}{3} \cdot 2 \cdot \int \frac{sec ( w )}{2} d w

Take the constant out:

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

= \frac{1}{3} \cdot 2 \cdot \frac{1}{2} \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= \frac{1}{3} \cdot 2 \cdot \frac{1}{2} ln ∣ tan (w) + sec (w) ∣

Substitute back

= \frac{1}{3} \cdot 2 \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{5} (3 x + 5 - \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (3 x + 5 - \frac{5}{2})))

Simplify

\frac{1}{3} \cdot 2 \cdot \frac{1}{2} ln tan (arcsec (\frac{2}{5} (3 x + 5 - \frac{5}{2}))) + sec (arcsec (\frac{2}{5} (3 x + 5 - \frac{5}{2}))) : \frac{1}{3} ln (\frac{36 x ^{2} + 60 x + 6 x + 5}{5})

= \frac{1}{3} ln (\frac{36 x ^{2} + 60 x + 6 x + 5}{5})

Add a constant to the solution

= \frac{1}{3} ln (\frac{36 x ^{2} + 60 x + 6 x + 5}{5}) + C

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What is the integral of 1/(sqrt(x(3x+5))) ?
The integral of 1/(sqrt(x(3x+5))) is 1/(sqrt(3))ln((|sqrt(36x^2+60x)+6x+5|)/5)+C