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

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

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

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

3 arctan (3 t) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

u = 3 t

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

Simplify

3 \cdot \frac{1}{3} arctan (3 t) : 3 arctan (3 t)

= 3 arctan (3 t)

Add a constant to the solution

= 3 arctan (3 t) + C

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