What is the integral from 6 to 10 of y/(y^2-2y-3) ?

The integral from 6 to 10 of y/(y^2-2y-3) is (ln(11)+2ln(7)-3ln(3))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 6 to 10 of y/(y^2-2y-3)

\int_{6}^{10} \frac{y}{y ^{2} - 2 y - 3} d y

\frac{ln ( 11 ) + 2 ln ( 7 ) - 3 ln ( 3 )}{4}

Decimal Notation

0.74846 \dots

Solution steps

\int_{6}^{10} \frac{y}{y ^{2} - 2 y - 3} d y

Take the partial fraction of

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

= \int_{6}^{10} \frac{1}{4 ( y + 1 )} + \frac{3}{4 ( y - 3 )} 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_{6}^{10} \frac{1}{4 ( y + 1 )} d y + \int_{6}^{10} \frac{3}{4 ( y - 3 )} d y

\int_{6}^{10} \frac{1}{4 ( y + 1 )} d y = \frac{1}{4} (ln (11) - ln (7))

\int_{6}^{10} \frac{3}{4 ( y - 3 )} d y = \frac{3}{4} (ln (7) - ln (3))

= \frac{1}{4} (ln (11) - ln (7)) + \frac{3}{4} (ln (7) - ln (3))

Simplify

\frac{1}{4} (ln (11) - ln (7)) + \frac{3}{4} (ln (7) - ln (3)) : \frac{ln ( 11 ) + 2 ln ( 7 ) - 3 ln ( 3 )}{4}

= \frac{ln ( 11 ) + 2 ln ( 7 ) - 3 ln ( 3 )}{4}

\int_{0}^{1} π (x^{2} + 1) d x

\int_{- 1}^{0} \frac{3 e ^{\frac{1}{x}}}{x ^{3}} d x

\int_{1}^{4} (x - (x - 2)^{2}) d x

\int_{\frac{π}{4}}^{\frac{3 π}{4}} 1

\int_{- π}^{π} sin (2 x) sin (x) d x

What is the integral from 6 to 10 of y/(y^2-2y-3) ?
The integral from 6 to 10 of y/(y^2-2y-3) is (ln(11)+2ln(7)-3ln(3))/4