What is the integral of 3/(sqrt(1+9y^2)) ?

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

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integral of 3/(sqrt(1+9y^2))

\int \frac{3}{1 + 9 y ^{2}} d y

ln 3 y + 1 + 9 y^{2} + C

Solution steps

\int \frac{3}{1 + 9 y ^{2}} d y

Take the constant out:

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

= 3 \cdot \int \frac{1}{1 + 9 y ^{2}} d y

Apply Trigonometric Substitution

= 3 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = arctan (3 y)

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

Simplify

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

= ln 3 y + 1 + 9 y^{2}

Add a constant to the solution

= ln 3 y + 1 + 9 y^{2} + C

\int 500 - \frac{x}{3} d x

\int \frac{- x}{2} d x

\int sin (1 + 2 x) d x

\int \frac{3}{( 2 x - 1 ) ^{2}} d x

\int cos (24 x) d x

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