What is the integral of (8t^3+10)/(4t^2-4t+5) ?

The integral of (8t^3+10)/(4t^2-4t+5) is t^2+2t-1/4 ln|4t^2-4t+5|-1/4 arctan(t-1/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8t^3+10)/(4t^2-4t+5)

\int \frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} d t

t^{2} + 2 t - \frac{1}{4} ln 4 t^{2} - 4 t + 5 - \frac{1}{4} arctan (t - \frac{1}{2}) + C

Solution steps

\int \frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} d t

Long division

\frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} : 2 t + 2 - \frac{2 t}{4 t ^{2} - 4 t + 5}

= \int 2 t + 2 - \frac{2 t}{4 t ^{2} - 4 t + 5} 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 2 t d t + \int 2 d t - \int \frac{2 t}{4 t ^{2} - 4 t + 5} d t

\int 2 t d t = t^{2}

\int 2 d t = 2 t

\int \frac{2 t}{4 t ^{2} - 4 t + 5} d t = \frac{1}{4} ln 4 t^{2} - 4 t + 5 + \frac{1}{4} arctan (t - \frac{1}{2})

= t^{2} + 2 t - (\frac{1}{4} ln 4 t^{2} - 4 t + 5 + \frac{1}{4} arctan (t - \frac{1}{2}))

Simplify

= t^{2} + 2 t - \frac{1}{4} ln 4 t^{2} - 4 t + 5 - \frac{1}{4} arctan (t - \frac{1}{2})

Add a constant to the solution

= t^{2} + 2 t - \frac{1}{4} ln 4 t^{2} - 4 t + 5 - \frac{1}{4} arctan (t - \frac{1}{2}) + C

\int x^{4} + 1 d x

\int \frac{3 x + 2}{19 - 5 x + x ^{2}} d x

x \to c lim (2.5 (- 2))

\int x^{2} (2 x - 5) d x

Q (x) = x^{4} + 8 x^{2} + 16

What is the integral of (8t^3+10)/(4t^2-4t+5) ?
The integral of (8t^3+10)/(4t^2-4t+5) is t^2+2t-1/4 ln|4t^2-4t+5|-1/4 arctan(t-1/2)+C