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

The integral from 2 to 3 of (2x)/((x^2+3)^3) is 95/14112

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{95}{14112}

Decimal

0.00673 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{7}^{12} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{7}^{12} \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{7}^{12} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{7}^{12}

= 2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{7}^{12}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{7}^{12} : [- \frac{1}{2 u ^{2}}]_{7}^{12}

= [- \frac{1}{2 u ^{2}}]_{7}^{12}

Compute the boundaries:

\frac{95}{14112}

= \frac{95}{14112}

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

\int_{0}^{x^{2}} f (t) d t

\int_{0}^{\infty} 0 d x

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

\int_{0}^{\frac{π}{2}} sin (x) \cdot cos (x) d x

What is the integral from 2 to 3 of (2x)/((x^2+3)^3) ?
The integral from 2 to 3 of (2x)/((x^2+3)^3) is 95/14112