What is the integral of 1/(sqrt(9y^2+6y+5)) ?

The integral of 1/(sqrt(9y^2+6y+5)) is 1/3 ln((|3y+1+sqrt(9y^2+6y+5)|)/2)+C

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integral of 1/(sqrt(9y^2+6y+5))

\int \frac{1}{9 y ^{2} + 6 y + 5} d y

\frac{1}{3} ln \frac{3 y + 1 + 9 y ^{2} + 6 y + 5}{2} + C

Solution steps

\int \frac{1}{9 y ^{2} + 6 y + 5} d y

Complete the square

9 y^{2} + 6 y + 5 : 9 (y + \frac{1}{3})^{2} + 4

= \int \frac{1}{9 ( y + \frac{1}{3} ) ^{2} + 4} d y

Apply u-substitution

: \int \frac{1}{9 u ^{2} + 4} d u

= \int \frac{1}{9 u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{3} d v

= \int \frac{sec ( v )}{3} d v

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{3} ln tan (arctan (\frac{3}{2} (y + \frac{1}{3}))) + sec (arctan (\frac{3}{2} (y + \frac{1}{3})))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{2} (y + \frac{1}{3}))) + sec (arctan (\frac{3}{2} (y + \frac{1}{3}))) : \frac{1}{3} ln \frac{3 y + 1 + 9 y ^{2} + 6 y + 5}{2}

= \frac{1}{3} ln \frac{3 y + 1 + 9 y ^{2} + 6 y + 5}{2}

Add a constant to the solution

= \frac{1}{3} ln \frac{3 y + 1 + 9 y ^{2} + 6 y + 5}{2} + C

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What is the integral of 1/(sqrt(9y^2+6y+5)) ?
The integral of 1/(sqrt(9y^2+6y+5)) is 1/3 ln((|3y+1+sqrt(9y^2+6y+5)|)/2)+C