What is the integral of (16s-4)/(3-2s+4s^2) ?

The integral of (16s-4)/(3-2s+4s^2) is 2ln|16s^2-8s+12|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (16s-4)/(3-2s+4s^2)

\int \frac{16 s - 4}{3 - 2 s + 4 s ^{2}} d s

2 ln 16 s^{2} - 8 s + 12 + C

Solution steps

\int \frac{16 s - 4}{3 - 2 s + 4 s ^{2}} d s

Expand

\frac{16 s - 4}{3 - 2 s + 4 s ^{2}} : \frac{16 s}{3 - 2 s + 4 s ^{2}} - \frac{4}{3 - 2 s + 4 s ^{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{16 s}{3 - 2 s + 4 s ^{2}} d s - \int \frac{4}{3 - 2 s + 4 s ^{2}} d s

\int \frac{16 s}{3 - 2 s + 4 s ^{2}} d s = 16 (\frac{1}{8} ln 16 s^{2} - 8 s + 12 + \frac{1}{4 11} arctan (\frac{4 s - 1}{11}))

\int \frac{4}{3 - 2 s + 4 s ^{2}} d s = \frac{4}{11} arctan (\frac{4 s - 1}{11})

= 16 (\frac{1}{8} ln 16 s^{2} - 8 s + 12 + \frac{1}{4 11} arctan (\frac{4 s - 1}{11})) - \frac{4}{11} arctan (\frac{4 s - 1}{11})

Simplify

16 (\frac{1}{8} ln 16 s^{2} - 8 s + 12 + \frac{1}{4 11} arctan (\frac{4 s - 1}{11})) - \frac{4}{11} arctan (\frac{4 s - 1}{11}) : 2 ln 16 s^{2} - 8 s + 12

= 2 ln 16 s^{2} - 8 s + 12

Add a constant to the solution

= 2 ln 16 s^{2} - 8 s + 12 + C

y = tan (e^{3 t}) + e^{t a n (3 t)}

f (x) = e^{r} + r^{e}

\int \frac{3 x}{x ^{2}} d x

f^{^{'}} (x) = 3 sec (x)

\int_{- 1}^{4} \frac{x}{5 + x} d x

What is the integral of (16s-4)/(3-2s+4s^2) ?
The integral of (16s-4)/(3-2s+4s^2) is 2ln|16s^2-8s+12|+C