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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

\int \frac{1 - x}{x} d x

\frac{10 lo g _{4} ( t )}{t}

\frac{\partial}{\partial x} (x^{2} sin (x y))

n = 0 \sum \infty 9^{n} x^{n}

9 x^{2} y^{^{'}} = y^{^{'}} + 9 x e^{- y}

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