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

The integral from-1 to 1 of (3x^2+1)^2 is 48/5

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

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

\frac{48}{5}

Decimal Notation

9.6

Solution steps

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

Expand

(3 x^{2} + 1)^{2} : 9 x^{4} + 6 x^{2} + 1

= \int_{- 1}^{1} 9 x^{4} + 6 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

= \int_{- 1}^{1} 9 x^{4} d x + \int_{- 1}^{1} 6 x^{2} d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} 9 x^{4} d x = \frac{18}{5}

\int_{- 1}^{1} 6 x^{2} d x = 4

\int_{- 1}^{1} 1 d x = 2

= \frac{18}{5} + 4 + 2

Simplify

\frac{18}{5} + 4 + 2 : \frac{48}{5}

= \frac{48}{5}

\int \frac{- 4}{x ^{2}} d x

y = \frac{2 x - 3}{4 x + 1}

\frac{\partial}{\partial x} (3 (x - y) \cdot e^{3 y + 2 x^{2}})

\frac{t}{1 + t ^{2}}

\int \frac{x ^{4}}{4 x ^{2} + 1} d x

What is the integral from-1 to 1 of (3x^2+1)^2 ?
The integral from-1 to 1 of (3x^2+1)^2 is 48/5