What is the integral of (x^2)/(16+x^2) ?

The integral of (x^2)/(16+x^2) is -4arctan(x/4)+x+C

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integral of (x^2)/(16+x^2)

\int \frac{x ^{2}}{16 + x ^{2}} d x

- 4 arctan (\frac{x}{4}) + x + C

Solution steps

\int \frac{x ^{2}}{16 + x ^{2}} d x

\frac{x ^{2}}{16 + x ^{2}} = - \frac{16}{16 + x ^{2}} + 1

= \int - \frac{16}{16 + x ^{2}} + 1 d x

Apply Integral Substitution

= \int 4 (- \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

= 4 \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

= 4 (- \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

= 4 (- arctan (u) + u)

Substitute back

u = \frac{x}{4}

= 4 (- arctan (\frac{x}{4}) + \frac{x}{4})

Simplify

4 (- arctan (\frac{x}{4}) + \frac{x}{4}) : - 4 arctan (\frac{x}{4}) + x

= - 4 arctan (\frac{x}{4}) + x

Add a constant to the solution

= - 4 arctan (\frac{x}{4}) + x + C

\int 4 t^{3} y - 15 t^{2} - y d t

11 x - 3, 10 x - 5, [12, 34]

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

x \to 129 lim (\frac{x - 129}{x + 15 - 12})

\frac{d}{d x} (3^{a r c t a n (x)})

What is the integral of (x^2)/(16+x^2) ?
The integral of (x^2)/(16+x^2) is -4arctan(x/4)+x+C