What is the integral of (4x^2+16x+16)/(x^3+8x^2+16x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^2+16x+16)/(x^3+8x^2+16x)

\int \frac{4 x ^{2} + 16 x + 16}{x ^{3} + 8 x ^{2} + 16 x} d x

3 ln ∣ x + 4 ∣ + \frac{4}{x + 4} + ln ∣ x ∣ + C

Solution steps

\int \frac{4 x ^{2} + 16 x + 16}{x ^{3} + 8 x ^{2} + 16 x} d x

Expand

\frac{4 x ^{2} + 16 x + 16}{x ^{3} + 8 x ^{2} + 16 x} : \frac{4 x ^{2}}{x ^{3} + 8 x ^{2} + 16 x} + \frac{16 x}{x ^{3} + 8 x ^{2} + 16 x} + \frac{16}{x ^{3} + 8 x ^{2} + 16 x}

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{4 x ^{2}}{x ^{3} + 8 x ^{2} + 16 x} d x + \int \frac{16 x}{x ^{3} + 8 x ^{2} + 16 x} d x + \int \frac{16}{x ^{3} + 8 x ^{2} + 16 x} d x

\int \frac{4 x ^{2}}{x ^{3} + 8 x ^{2} + 16 x} d x = 4 (ln ∣ x + 4 ∣ + \frac{4}{x + 4})

\int \frac{16 x}{x ^{3} + 8 x ^{2} + 16 x} d x = - \frac{16}{x + 4}

\int \frac{16}{x ^{3} + 8 x ^{2} + 16 x} d x = ln ∣ x ∣ - ln ∣ x + 4 ∣ + \frac{4}{x + 4}

= 4 (ln ∣ x + 4 ∣ + \frac{4}{x + 4}) - \frac{16}{x + 4} + ln ∣ x ∣ - ln ∣ x + 4 ∣ + \frac{4}{x + 4}

Simplify

4 (ln ∣ x + 4 ∣ + \frac{4}{x + 4}) - \frac{16}{x + 4} + ln ∣ x ∣ - ln ∣ x + 4 ∣ + \frac{4}{x + 4} : 3 ln ∣ x + 4 ∣ + \frac{4}{x + 4} + ln ∣ x ∣

= 3 ln ∣ x + 4 ∣ + \frac{4}{x + 4} + ln ∣ x ∣

Add a constant to the solution

= 3 ln ∣ x + 4 ∣ + \frac{4}{x + 4} + ln ∣ x ∣ + C

\int \frac{x}{( 5 - 4 x ) ^{\frac{3}{2}}} d x

\int e^{(\frac{- x ^{2}}{2})} d x

\int \frac{1}{( 2 - y ) ^{3}} d y

\int \frac{18}{x ^{3} + x ^{2}} d x

\int 7 (\frac{sin ( 2 x )}{sin ( x )}) d x

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