What is the integral of (x^2)/(4+x^2) ?

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

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integral of (x^2)/(4+x^2)

\int \frac{x ^{2}}{4 + x ^{2}} d x

- 2 arctan (\frac{x}{2}) + x + C

Solution steps

\int \frac{x ^{2}}{4 + x ^{2}} d x

\frac{x ^{2}}{4 + x ^{2}} = - \frac{4}{4 + x ^{2}} + 1

= \int - \frac{4}{4 + x ^{2}} + 1 d x

Apply Integral Substitution

= \int 2 (- \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

= 2 \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

= 2 (- \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

= 2 (- arctan (u) + u)

Substitute back

u = \frac{x}{2}

= 2 (- arctan (\frac{x}{2}) + \frac{x}{2})

Simplify

2 (- arctan (\frac{x}{2}) + \frac{x}{2}) : - 2 arctan (\frac{x}{2}) + x

= - 2 arctan (\frac{x}{2}) + x

Add a constant to the solution

= - 2 arctan (\frac{x}{2}) + x + C

\frac{\partial}{\partial x} (y cos (π x^{2} y))

f (x) = cos (e^{x})

x \to - 2 lim (\frac{3}{2})

\frac{d y}{d x} = 0.1 ln (\frac{3000}{y}) y

\frac{\partial}{\partial x} (2 x^{2} \cdot 4 x^{3})

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