What is the integral from 7 to 10 of y/(y^2-3y-4) ?

The integral from 7 to 10 of y/(y^2-3y-4) is (ln(11)+ln(2))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 7 to 10 of y/(y^2-3y-4)

\int_{7}^{10} \frac{y}{y ^{2} - 3 y - 4} d y

\frac{ln ( 11 ) + ln ( 2 )}{5}

Decimal

0.61820 \dots

Solution steps

\int_{7}^{10} \frac{y}{y ^{2} - 3 y - 4} d y

Take the partial fraction of

\frac{y}{y ^{2} - 3 y - 4} : \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )}

= \int_{7}^{10} \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{7}^{10} \frac{1}{5 ( y + 1 )} d y + \int_{7}^{10} \frac{4}{5 ( y - 4 )} d y

\int_{7}^{10} \frac{1}{5 ( y + 1 )} d y = \frac{1}{5} (ln (11) - 3 ln (2))

\int_{7}^{10} \frac{4}{5 ( y - 4 )} d y = \frac{4}{5} ln (2)

= \frac{1}{5} (ln (11) - 3 ln (2)) + \frac{4}{5} ln (2)

Simplify

\frac{1}{5} (ln (11) - 3 ln (2)) + \frac{4}{5} ln (2) : \frac{ln ( 11 ) + ln ( 2 )}{5}

= \frac{ln ( 11 ) + ln ( 2 )}{5}

What is the integral from 7 to 10 of y/(y^2-3y-4) ?
The integral from 7 to 10 of y/(y^2-3y-4) is (ln(11)+ln(2))/5