What is the integral of 8/(t^2(t^2-4)) ?

The integral of 8/(t^2(t^2-4)) is 8(1/(4t)-1/16 ln|t+2|+1/16 ln|t-2|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(t^2(t^2-4))

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

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

Solution steps

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

Take the constant out:

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

= 8 \cdot \int \frac{1}{t ^{2} ( t ^{2} - 4 )} d t

Take the partial fraction of

\frac{1}{t ^{2} ( t ^{2} - 4 )} : - \frac{1}{4 t ^{2}} - \frac{1}{16 ( t + 2 )} + \frac{1}{16 ( t - 2 )}

= 8 \cdot \int - \frac{1}{4 t ^{2}} - \frac{1}{16 ( t + 2 )} + \frac{1}{16 ( t - 2 )} 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

= 8 (- \int \frac{1}{4 t ^{2}} d t - \int \frac{1}{16 ( t + 2 )} d t + \int \frac{1}{16 ( t - 2 )} d t)

\int \frac{1}{4 t ^{2}} d t = - \frac{1}{4 t}

\int \frac{1}{16 ( t + 2 )} d t = \frac{1}{16} ln ∣ t + 2 ∣

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

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

Simplify

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

Add a constant to the solution

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

x \to 2 lim (5 x^{2} - 4 x + 3)

y = 9 x - 4 x^{2}, (- 2, - 34)

\int \frac{cos ( \frac{π}{x ^{5}} )}{x ^{6}} d x

\int sin (ln (x)) d x

\int 0.5 x d x

What is the integral of 8/(t^2(t^2-4)) ?
The integral of 8/(t^2(t^2-4)) is 8(1/(4t)-1/16 ln|t+2|+1/16 ln|t-2|)+C