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

The integral of 1/((2y+1)sqrt((2y+1)^2-9)) is 1/6 arcsec(1/3 (2y+1))+C

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

\int \frac{1}{( 2 y + 1 ) ( 2 y + 1 ) ^{2} - 9} d y

\frac{1}{6} arcsec (\frac{1}{3} (2 y + 1)) + C

Solution steps

\int \frac{1}{( 2 y + 1 ) ( 2 y + 1 ) ^{2} - 9} d y

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u u ^{2} - 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \frac{1}{2} \cdot \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (2 y + 1))

Simplify

\frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (2 y + 1)) : \frac{1}{6} arcsec (\frac{1}{3} (2 y + 1))

= \frac{1}{6} arcsec (\frac{1}{3} (2 y + 1))

Add a constant to the solution

= \frac{1}{6} arcsec (\frac{1}{3} (2 y + 1)) + C

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