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

The integral from 3 to 5 of 5/((2x+1)(x-2)) is -ln(11)+ln(7)+ln(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- ln (11) + ln (7) + ln (3)

Decimal Notation

0.64662 \dots

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{( 2 x + 1 ) ( x - 2 )} : - \frac{2}{5 ( 2 x + 1 )} + \frac{1}{5 ( x - 2 )}

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

\int_{3}^{5} \frac{2}{5 ( 2 x + 1 )} d x = \frac{1}{5} (ln (11) - ln (7))

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

= 5 (- \frac{1}{5} (ln (11) - ln (7)) + \frac{1}{5} ln (3))

Simplify

5 (- \frac{1}{5} (ln (11) - ln (7)) + \frac{1}{5} ln (3)) : - ln (11) + ln (7) + ln (3)

= - ln (11) + ln (7) + ln (3)

\int_{0}^{0.4} cos (x) + ln (x + 1) d x

\int_{1}^{e} (ln (x))^{5} d x

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

\int_{0}^{4} (3 x + 4)^{\frac{3}{2}} d x

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

What is the integral from 3 to 5 of 5/((2x+1)(x-2)) ?
The integral from 3 to 5 of 5/((2x+1)(x-2)) is -ln(11)+ln(7)+ln(3)