What is the integral of 1/(sqrt(9y^2+64)) ?

The integral of 1/(sqrt(9y^2+64)) is 1/3 ln((|3y+sqrt(64+9y^2)|)/8)+C

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integral of 1/(sqrt(9y^2+64))

\int \frac{1}{9 y ^{2} + 64} d y

\frac{1}{3} ln \frac{3 y + 64 + 9 y ^{2}}{8} + C

Solution steps

\int \frac{1}{9 y ^{2} + 64} d y

Apply Trigonometric Substitution

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

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

Use the common integral:

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

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

Substitute back

u = arctan (\frac{3}{8} y)

= \frac{1}{3} ln tan (arctan (\frac{3}{8} y)) + sec (arctan (\frac{3}{8} y))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{8} y)) + sec (arctan (\frac{3}{8} y)) : \frac{1}{3} ln \frac{3 y + 64 + 9 y ^{2}}{8}

= \frac{1}{3} ln \frac{3 y + 64 + 9 y ^{2}}{8}

Add a constant to the solution

= \frac{1}{3} ln \frac{3 y + 64 + 9 y ^{2}}{8} + C

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What is the integral of 1/(sqrt(9y^2+64)) ?
The integral of 1/(sqrt(9y^2+64)) is 1/3 ln((|3y+sqrt(64+9y^2)|)/8)+C