What is the integral of (2t^2-t+4)/(t^3+4t) ?

The integral of (2t^2-t+4)/(t^3+4t) is ln|t|+1/2 ln|t^2+4|-1/2 arctan(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 (2t^2-t+4)/(t^3+4t)

\int \frac{2 t ^{2} - t + 4}{t ^{3} + 4 t} d t

ln ∣ t ∣ + \frac{1}{2} ln t^{2} + 4 - \frac{1}{2} arctan (\frac{t}{2}) + C

Solution steps

\int \frac{2 t ^{2} - t + 4}{t ^{3} + 4 t} d t

Take the partial fraction of

\frac{2 t ^{2} - t + 4}{t ^{3} + 4 t} : \frac{1}{t} + \frac{t - 1}{t ^{2} + 4}

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

\int \frac{1}{t} d t = ln ∣ t ∣

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

= ln ∣ t ∣ + \frac{1}{2} ln t^{2} + 4 - \frac{1}{2} arctan (\frac{t}{2})

Add a constant to the solution

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

\int 6 tan^{2} (x) sec^{4} (x) d x

\frac{1}{( x ^{2} + x x )}

\int (2 x^{2} + 3 x + 2) d x

\int (2 x - 1)^{- 1} d x

\int \frac{5 + x}{cos ^{2} ( x )} d x

What is the integral of (2t^2-t+4)/(t^3+4t) ?
The integral of (2t^2-t+4)/(t^3+4t) is ln|t|+1/2 ln|t^2+4|-1/2 arctan(t/2)+C