What is the integral from 0 to 1 of 3x^2(x^3+5)^{-2} ?

The integral from 0 to 1 of 3x^2(x^3+5)^{-2} is 1/30

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 3x^2(x^3+5)^{-2}

\int_{0}^{1} 3 x^{2} (x^{3} + 5)^{- 2} d x

\frac{1}{30}

Decimal Notation

0.03333 \dots

Solution steps

\int_{0}^{1} 3 x^{2} (x^{3} + 5)^{- 2} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} x^{2} (x^{3} + 5)^{- 2} d x

Simplify

x^{2} (x^{3} + 5)^{- 2} : \frac{x ^{2}}{( x ^{3} + 5 ) ^{2}}

= 3 \cdot \int_{0}^{1} \frac{x ^{2}}{( x ^{3} + 5 ) ^{2}} d x

Apply u-substitution

: \int_{5}^{6} \frac{1}{3 u ^{2}} d u

= 3 \cdot \int_{5}^{6} \frac{1}{3 u ^{2}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{5}^{6} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{5}^{6}

= 3 \cdot \frac{1}{3} [- \frac{1}{u}]_{5}^{6}

Simplify

3 \cdot \frac{1}{3} [- \frac{1}{u}]_{5}^{6} : [- \frac{1}{u}]_{5}^{6}

= [- \frac{1}{u}]_{5}^{6}

Compute the boundaries:

\frac{1}{30}

= \frac{1}{30}

\int sin (5 π) t d t

2 y^{^{''}} - 8 y = 0

\frac{d}{d x} (- e^{x} (x - 7))

\frac{\partial}{\partial y} (- 10 y sin (x y))

y = x^{x^{2}}

What is the integral from 0 to 1 of 3x^2(x^3+5)^{-2} ?
The integral from 0 to 1 of 3x^2(x^3+5)^{-2} is 1/30