What is the integral of (t^4+1)/(t^5+4t^3) ?

The integral of (t^4+1)/(t^5+4t^3) is -1/16 ln|t|-1/(8t^2)+17/32 ln|t^2+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^4+1)/(t^5+4t^3)

\int \frac{t ^{4} + 1}{t ^{5} + 4 t ^{3}} d t

- \frac{1}{16} ln ∣ t ∣ - \frac{1}{8 t ^{2}} + \frac{17}{32} ln t^{2} + 4 + C

Solution steps

\int \frac{t ^{4} + 1}{t ^{5} + 4 t ^{3}} d t

Take the partial fraction of

\frac{t ^{4} + 1}{t ^{5} + 4 t ^{3}} : - \frac{1}{16 t} + \frac{1}{4 t ^{3}} + \frac{17 t}{16 ( t ^{2} + 4 )}

= \int - \frac{1}{16 t} + \frac{1}{4 t ^{3}} + \frac{17 t}{16 ( t ^{2} + 4 )} 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{1}{16 t} d t + \int \frac{1}{4 t ^{3}} d t + \int \frac{17 t}{16 ( t ^{2} + 4 )} d t

\int \frac{1}{16 t} d t = \frac{1}{16} ln ∣ t ∣

\int \frac{1}{4 t ^{3}} d t = - \frac{1}{8 t ^{2}}

\int \frac{17 t}{16 ( t ^{2} + 4 )} d t = \frac{17}{32} ln t^{2} + 4

= - \frac{1}{16} ln ∣ t ∣ - \frac{1}{8 t ^{2}} + \frac{17}{32} ln t^{2} + 4

Add a constant to the solution

= - \frac{1}{16} ln ∣ t ∣ - \frac{1}{8 t ^{2}} + \frac{17}{32} ln t^{2} + 4 + C

(x^{2} + 1) y^{^{'}} + 9 x (y - 1) = 0, y (0) = 4

\int u^{2} u^{3} + 2 d u

sin (x^{7})

y + 2 t^{2} + y^{^{'}} t ln (t) = 0

\int \frac{x + 8}{x ^{2} + 16} d x

What is the integral of (t^4+1)/(t^5+4t^3) ?
The integral of (t^4+1)/(t^5+4t^3) is -1/16 ln|t|-1/(8t^2)+17/32 ln|t^2+4|+C