What is the integral of (6t-8)/((t-3)^3) ?

The integral of (6t-8)/((t-3)^3) is -6/(t-3)-5/((t-3)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (6t-8)/((t-3)^3)

\int \frac{6 t - 8}{( t - 3 ) ^{3}} d t

- \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}} + C

Solution steps

\int \frac{6 t - 8}{( t - 3 ) ^{3}} d t

Expand

\frac{6 t - 8}{( t - 3 ) ^{3}} : \frac{6 t}{( t - 3 ) ^{3}} - \frac{8}{( t - 3 ) ^{3}}

= \int \frac{6 t}{( t - 3 ) ^{3}} - \frac{8}{( t - 3 ) ^{3}} d t

Apply the Sum Rule:

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

= \int \frac{6 t}{( t - 3 ) ^{3}} d t - \int \frac{8}{( t - 3 ) ^{3}} d t

\int \frac{6 t}{( t - 3 ) ^{3}} d t = 6 (- \frac{1}{t - 3} - \frac{3}{2 ( t - 3 ) ^{2}})

\int \frac{8}{( t - 3 ) ^{3}} d t = - \frac{4}{( t - 3 ) ^{2}}

= 6 (- \frac{1}{t - 3} - \frac{3}{2 ( t - 3 ) ^{2}}) - (- \frac{4}{( t - 3 ) ^{2}})

Simplify

6 (- \frac{1}{t - 3} - \frac{3}{2 ( t - 3 ) ^{2}}) - (- \frac{4}{( t - 3 ) ^{2}}) : - \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}}

= - \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}}

Add a constant to the solution

= - \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}} + C

What is the integral of (6t-8)/((t-3)^3) ?
The integral of (6t-8)/((t-3)^3) is -6/(t-3)-5/((t-3)^2)+C