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

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

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

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

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

Solution steps

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

\frac{x ^{2}}{36 + x ^{2}} = - \frac{36}{36 + x ^{2}} + 1

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

Apply Integral Substitution

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

Take the constant out:

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

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

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

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

\int 1 d u = u

= 6 (- arctan (u) + u)

Substitute back

u = \frac{x}{6}

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

Simplify

6 (- arctan (\frac{x}{6}) + \frac{x}{6}) : - 6 arctan (\frac{x}{6}) + x

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

Add a constant to the solution

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

\frac{1}{6 x + 7}

\frac{d}{d x} (cos (x) ln (x))

\frac{d}{d x} (\frac{sin ( 6 ) x ^{3}}{5 x})

\int x (ln (x + 1)) d x

\frac{d}{d y} (e^{x y})

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