What is the integral of 1/(x^3+3x^2+7x+5) ?

The integral of 1/(x^3+3x^2+7x+5) is 1/4 ln|x+1|-1/8 ln|x^2+2x+5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^3+3x^2+7x+5)

\int \frac{1}{x ^{3} + 3 x ^{2} + 7 x + 5} d x

\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 2 x + 5 + C

Solution steps

\int \frac{1}{x ^{3} + 3 x ^{2} + 7 x + 5} d x

Take the partial fraction of

\frac{1}{x ^{3} + 3 x ^{2} + 7 x + 5} : \frac{1}{4 ( x + 1 )} + \frac{- x - 1}{4 ( x ^{2} + 2 x + 5 )}

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{- x - 1}{4 ( x ^{2} + 2 x + 5 )} d x = - \frac{1}{8} ln x^{2} + 2 x + 5

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 2 x + 5

Add a constant to the solution

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 2 x + 5 + C

\int_{0}^{1} 2 π (x^{5} - x^{6}) d x

(x y^{3} + x^{2} y)^{3}

x \to 4 lim (f (x))

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

f (x) = - 4 x

What is the integral of 1/(x^3+3x^2+7x+5) ?
The integral of 1/(x^3+3x^2+7x+5) is 1/4 ln|x+1|-1/8 ln|x^2+2x+5|+C