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

The integral from 1 to 3 of (2x)/((x^2+3)^3) is 1/36

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{36}

Decimal

0.02777 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{12} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{4}^{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_{4}^{12} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{4}^{12}

= 2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{4}^{12}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{4}^{12} : [- \frac{1}{2 u ^{2}}]_{4}^{12}

= [- \frac{1}{2 u ^{2}}]_{4}^{12}

Compute the boundaries:

\frac{1}{36}

= \frac{1}{36}

\int_{2}^{e} \frac{x ^{4} - 5 x}{x ^{2}} d x

\int_{0}^{1} x^{- \frac{1}{2}} ln (x) d x

\int_{- \frac{π}{3}}^{\frac{π}{3}} sin (x) d x

\int_{1}^{3} 4 x (x^{2} + 5) d x

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

What is the integral from 1 to 3 of (2x)/((x^2+3)^3) ?
The integral from 1 to 3 of (2x)/((x^2+3)^3) is 1/36