What is the integral of 6/(1+6t^2) ?

The integral of 6/(1+6t^2) is sqrt(6)arctan(sqrt(6)t)+C

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integral of 6/(1+6t^2)

\int \frac{6}{1 + 6 t ^{2}} d t

6 arctan (6 t) + C

Solution steps

\int \frac{6}{1 + 6 t ^{2}} d t

Take the constant out:

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

= 6 \cdot \int \frac{1}{1 + 6 t ^{2}} d t

Apply Integral Substitution

= 6 \cdot \int \frac{1}{6 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} arctan (u)

Substitute back

u = 6 t

= 6 \cdot \frac{1}{6} arctan (6 t)

Simplify

6 \cdot \frac{1}{6} arctan (6 t) : 6 arctan (6 t)

= 6 arctan (6 t)

Add a constant to the solution

= 6 arctan (6 t) + C

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What is the integral of 6/(1+6t^2) ?
The integral of 6/(1+6t^2) is sqrt(6)arctan(sqrt(6)t)+C