What is the integral of (t-2)/(t^2(t-4)^2) ?

The integral of (t-2)/(t^2(t-4)^2) is -1/(4(t-4))+1/(8t)+1/(8(t-4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (t-2)/(t^2(t-4)^2)

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

- \frac{1}{4 ( t - 4 )} + \frac{1}{8 t} + \frac{1}{8 ( t - 4 )} + C

Solution steps

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

Expand

\frac{t - 2}{t ^{2} ( t - 4 ) ^{2}} : \frac{t}{t ^{2} ( t - 4 ) ^{2}} - \frac{2}{t ^{2} ( t - 4 ) ^{2}}

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

\int \frac{t}{t ^{2} ( t - 4 ) ^{2}} d t = \frac{1}{16} ln ∣ t ∣ - \frac{1}{16} ln ∣ t - 4 ∣ - \frac{1}{4 ( t - 4 )}

\int \frac{2}{t ^{2} ( t - 4 ) ^{2}} d t = 2 (\frac{1}{32} ln ∣ t ∣ - \frac{1}{16 t} - \frac{1}{32} ln ∣ t - 4 ∣ - \frac{1}{16 ( t - 4 )})

= \frac{1}{16} ln ∣ t ∣ - \frac{1}{16} ln ∣ t - 4 ∣ - \frac{1}{4 ( t - 4 )} - 2 (\frac{1}{32} ln ∣ t ∣ - \frac{1}{16 t} - \frac{1}{32} ln ∣ t - 4 ∣ - \frac{1}{16 ( t - 4 )})

Simplify

\frac{1}{16} ln ∣ t ∣ - \frac{1}{16} ln ∣ t - 4 ∣ - \frac{1}{4 ( t - 4 )} - 2 (\frac{1}{32} ln ∣ t ∣ - \frac{1}{16 t} - \frac{1}{32} ln ∣ t - 4 ∣ - \frac{1}{16 ( t - 4 )}) : - \frac{1}{4 ( t - 4 )} + \frac{1}{8 t} + \frac{1}{8 ( t - 4 )}

= - \frac{1}{4 ( t - 4 )} + \frac{1}{8 t} + \frac{1}{8 ( t - 4 )}

Add a constant to the solution

= - \frac{1}{4 ( t - 4 )} + \frac{1}{8 t} + \frac{1}{8 ( t - 4 )} + C

\int x cosh (a) x d x

\int (t + t^{2}) e^{- t} d t

\int sec^{2} (t) tan^{2} (t) d t

\int \frac{3 x - 1}{x ^{2} - 5 x} d x

\int \frac{( x + 1 )}{x ^{2} + 2 x + 4} d x

What is the integral of (t-2)/(t^2(t-4)^2) ?
The integral of (t-2)/(t^2(t-4)^2) is -1/(4(t-4))+1/(8t)+1/(8(t-4))+C