What is the integral from 1 to 2 of 5/(x^3+4x) ?

The integral from 1 to 2 of 5/(x^3+4x) is 5*(-ln(2)+ln(5))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of 5/(x^3+4x)

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

5 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

Decimal Notation

0.57268 \dots

Solution steps

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

Take the constant out:

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

= 5 \cdot \int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 4 x} : \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

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

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

\int_{1}^{2} \frac{1}{4 x} d x = \frac{1}{4} ln (2)

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= 5 (\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)))

Join

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)) : \frac{- ln ( 2 ) + ln ( 5 )}{8}

= 5 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

x \to 0 + lim (\frac{1}{x ^{3}})

(arcsin (3 x))^{^{'}}

6 x e^{x} csc (x)

\int \frac{2}{x} d x

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

What is the integral from 1 to 2 of 5/(x^3+4x) ?
The integral from 1 to 2 of 5/(x^3+4x) is 5*(-ln(2)+ln(5))/8