What is the integral from 0 to 1 of (x^2)/(25+x^2) ?

The integral from 0 to 1 of (x^2)/(25+x^2) is -5arctan(1/5)+1

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integral from 0 to 1 of (x^2)/(25+x^2)

\int_{0}^{1} \frac{x ^{2}}{25 + x ^{2}} d x

- 5 arctan (\frac{1}{5}) + 1

Decimal

0.01302 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{25 + x ^{2}} d x

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

= \int_{0}^{1} - \frac{25}{25 + x ^{2}} + 1 d x

Apply Integral Substitution

= \int_{0}^{\frac{1}{5}} 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_{0}^{\frac{1}{5}} - \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_{0}^{\frac{1}{5}} \frac{1}{1 + u ^{2}} d u + \int_{0}^{\frac{1}{5}} 1 d u)

\int_{0}^{\frac{1}{5}} \frac{1}{1 + u ^{2}} d u = arctan (\frac{1}{5})

\int_{0}^{\frac{1}{5}} 1 d u = \frac{1}{5}

= 5 (- arctan (\frac{1}{5}) + \frac{1}{5})

Simplify

5 (- arctan (\frac{1}{5}) + \frac{1}{5}) : - 5 arctan (\frac{1}{5}) + 1

= - 5 arctan (\frac{1}{5}) + 1

What is the integral from 0 to 1 of (x^2)/(25+x^2) ?
The integral from 0 to 1 of (x^2)/(25+x^2) is -5arctan(1/5)+1