What is the integral from 0 to 1 of ((4x^2))/(1+x^2) ?

The integral from 0 to 1 of ((4x^2))/(1+x^2) is -pi+4

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integral from 0 to 1 of ((4x^2))/(1+x^2)

\int_{0}^{1} \frac{( 4 x ^{2} )}{1 + x ^{2}} d x

- π + 4

Decimal Notation

0.85840 \dots

Solution steps

\int_{0}^{1} \frac{4 x ^{2}}{1 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{1} \frac{x ^{2}}{1 + x ^{2}} d x

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

= 4 \cdot \int_{0}^{1} - \frac{1}{1 + x ^{2}} + 1 d x

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

\int_{0}^{1} \frac{1}{1 + x ^{2}} d x = \frac{π}{4}

\int_{0}^{1} 1 d x = 1

= 4 (- \frac{π}{4} + 1)

Simplify

4 (- \frac{π}{4} + 1) : - π + 4

= - π + 4

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What is the integral from 0 to 1 of ((4x^2))/(1+x^2) ?
The integral from 0 to 1 of ((4x^2))/(1+x^2) is -pi+4