What is the integral of 1/(sqrt(9t^2+4)) ?

The integral of 1/(sqrt(9t^2+4)) is 1/3 ln((|3t+sqrt(4+9t^2)|)/2)+C

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integral of 1/(sqrt(9t^2+4))

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

\frac{1}{3} ln (\frac{3 t + 4 + 9 t ^{2}}{2}) + C

Solution steps

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

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}{2} t)

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

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{2} t)) + sec (arctan (\frac{3}{2} t)) : \frac{1}{3} ln (\frac{3 t + 4 + 9 t ^{2}}{2})

= \frac{1}{3} ln (\frac{3 t + 4 + 9 t ^{2}}{2})

Add a constant to the solution

= \frac{1}{3} ln (\frac{3 t + 4 + 9 t ^{2}}{2}) + C

\int \frac{3 sin ( x ) + 2}{2 sin ( x ) + 3} d x

\int \frac{2}{( x - \frac{1}{x} ) x ^{4} - 2 x ^{2} - 8} d x

\int \frac{5 x ^{2}}{( 9 + x ^{2} ) ^{2}} d x

\int \frac{1}{x ^{2} + b ^{4}} d x

What is the integral of 1/(sqrt(9t^2+4)) ?
The integral of 1/(sqrt(9t^2+4)) is 1/3 ln((|3t+sqrt(4+9t^2)|)/2)+C