What is the integral from 0 to 1 of (5x)/((x^2+4)^2) ?

The integral from 0 to 1 of (5x)/((x^2+4)^2) is 1/8

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 (5x)/((x^2+4)^2)

\int_{0}^{1} \frac{5 x}{( x ^{2} + 4 ) ^{2}} d x

\frac{1}{8}

Decimal Notation

0.125

Solution steps

\int_{0}^{1} \frac{5 x}{( x ^{2} + 4 ) ^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{x}{( x ^{2} + 4 ) ^{2}} d x

Apply u-substitution

: \int_{4}^{5} \frac{1}{2 u ^{2}} d u

= 5 \cdot \int_{4}^{5} \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int_{4}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{4}^{5}

= 5 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{5}

Simplify

5 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{5} : \frac{5}{2} [- \frac{1}{u}]_{4}^{5}

= \frac{5}{2} [- \frac{1}{u}]_{4}^{5}

Compute the boundaries:

\frac{1}{20}

= \frac{5}{2} \cdot \frac{1}{20}

Simplify

= \frac{1}{8}

\int_{- 1}^{1} x^{3} e^{x^{4}} d x

\int_{3.3}^{8.3} (12 x^{2} - 6 x + 4) d x

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

\int_{2}^{8} x^{2} + 1 d x

\int_{0}^{\frac{π}{4}} sin^{2} (t) d t

What is the integral from 0 to 1 of (5x)/((x^2+4)^2) ?
The integral from 0 to 1 of (5x)/((x^2+4)^2) is 1/8