What is the integral of (3x^2)/(36x^2+1) ?

The integral of (3x^2)/(36x^2+1) is 1/72 (-arctan(6x)+6x)+C

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integral of (3x^2)/(36x^2+1)

\int \frac{3 x ^{2}}{36 x ^{2} + 1} d x

\frac{1}{72} (- arctan (6 x) + 6 x) + C

Solution steps

\int \frac{3 x ^{2}}{36 x ^{2} + 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{2}}{36 x ^{2} + 1} d x

Apply Integral Substitution

= 3 \cdot \int \frac{u ^{2}}{216 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{216} \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= 3 \cdot \frac{1}{216} \cdot \int - \frac{1}{u ^{2} + 1} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{216} (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

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

\int 1 d u = u

= 3 \cdot \frac{1}{216} (- arctan (u) + u)

Substitute back

u = 6 x

= 3 \cdot \frac{1}{216} (- arctan (6 x) + 6 x)

Simplify

3 \cdot \frac{1}{216} (- arctan (6 x) + 6 x) : \frac{1}{72} (- arctan (6 x) + 6 x)

= \frac{1}{72} (- arctan (6 x) + 6 x)

Add a constant to the solution

= \frac{1}{72} (- arctan (6 x) + 6 x) + C

What is the integral of (3x^2)/(36x^2+1) ?
The integral of (3x^2)/(36x^2+1) is 1/72 (-arctan(6x)+6x)+C