What is the integral from 0 to 1 of (x^2)/(3+x^2) ?

The integral from 0 to 1 of (x^2)/(3+x^2) is sqrt(3)(-pi/6+1/(sqrt(3)))

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integral from 0 to 1 of (x^2)/(3+x^2)

\int_{0}^{1} \frac{x ^{2}}{3 + x ^{2}} d x

3 (- \frac{π}{6} + \frac{1}{3})

Decimal

0.09310 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{3 + x ^{2}} d x

\frac{x ^{2}}{3 + x ^{2}} = - \frac{3}{3 + x ^{2}} + 1

= \int_{0}^{1} - \frac{3}{3 + x ^{2}} + 1 d x

Apply Integral Substitution

= \int_{0}^{\frac{1}{3}} 3 (- \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

= 3 \cdot \int_{0}^{\frac{1}{3}} - \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

= 3 (- \int_{0}^{\frac{1}{3}} \frac{1}{1 + u ^{2}} d u + \int_{0}^{\frac{1}{3}} 1 d u)

\int_{0}^{\frac{1}{3}} \frac{1}{1 + u ^{2}} d u = \frac{π}{6}

\int_{0}^{\frac{1}{3}} 1 d u = \frac{1}{3}

= 3 (- \frac{π}{6} + \frac{1}{3})

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What is the integral from 0 to 1 of (x^2)/(3+x^2) ?
The integral from 0 to 1 of (x^2)/(3+x^2) is sqrt(3)(-pi/6+1/(sqrt(3)))