What is the integral from-1 to 0 of (3x^4)/(x^2+1) ?

The integral from-1 to 0 of (3x^4)/(x^2+1) is -2+3*pi/4

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integral from-1 to 0 of (3x^4)/(x^2+1)

\int_{- 1}^{0} \frac{3 x ^{4}}{x ^{2} + 1} d x

- 2 + 3 \cdot \frac{π}{4}

Solution steps

\int_{- 1}^{0} \frac{3 x ^{4}}{x ^{2} + 1} d x

Take the constant out:

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

= 3 \cdot \int_{- 1}^{0} \frac{x ^{4}}{x ^{2} + 1} d x

Long division

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

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

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

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

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

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

= 3 (\frac{1}{3} - 1 + \frac{π}{4})

Simplify

3 (\frac{1}{3} - 1 + \frac{π}{4}) : - 2 + 3 \cdot \frac{π}{4}

= - 2 + 3 \cdot \frac{π}{4}

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What is the integral from-1 to 0 of (3x^4)/(x^2+1) ?
The integral from-1 to 0 of (3x^4)/(x^2+1) is -2+3*pi/4