What is the integral of 1/(4+3t^2) ?

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

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integral of 1/(4+3t^2)

\int \frac{1}{4 + 3 t ^{2}} d t

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

Solution steps

\int \frac{1}{4 + 3 t ^{2}} d t

Apply Integral Substitution

= \int \frac{1}{2 3 ( u ^{2} + 1 )} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2 3} arctan (u)

Substitute back

u = \frac{3}{2} t

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

Add a constant to the solution

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

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