What is the integral of (x^2)/(25+x^2) ?

The integral of (x^2)/(25+x^2) is -5arctan(x/5)+x+C

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integral of (x^2)/(25+x^2)

\int \frac{x ^{2}}{25 + x ^{2}} d x

- 5 arctan (\frac{x}{5}) + x + C

Solution steps

\int \frac{x ^{2}}{25 + x ^{2}} d x

\frac{x ^{2}}{25 + x ^{2}} = - \frac{25}{25 + x ^{2}} + 1

= \int - \frac{25}{25 + x ^{2}} + 1 d x

Apply Integral Substitution

= \int 5 (- \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

= 5 \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

= 5 (- \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

= 5 (- arctan (u) + u)

Substitute back

u = \frac{x}{5}

= 5 (- arctan (\frac{x}{5}) + \frac{x}{5})

Simplify

5 (- arctan (\frac{x}{5}) + \frac{x}{5}) : - 5 arctan (\frac{x}{5}) + x

= - 5 arctan (\frac{x}{5}) + x

Add a constant to the solution

= - 5 arctan (\frac{x}{5}) + x + C

\frac{\partial}{\partial x} (4 y (\frac{1}{x} - ln (x)))

\int_{0}^{π} x sin (2 x) d x

\frac{\partial}{\partial x} (x e^{15 x y})

\frac{\partial}{\partial x} (- y + x + 2)

What is the integral of (x^2)/(25+x^2) ?
The integral of (x^2)/(25+x^2) is -5arctan(x/5)+x+C