What is the integral from 2 to 3 of 2/(x^3+x^2) ?

The integral from 2 to 3 of 2/(x^3+x^2) is 2(3ln(2)+1/6-2ln(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 3 of 2/(x^3+x^2)

\int_{2}^{3} \frac{2}{x ^{3} + x ^{2}} d x

2 (3 ln (2) + \frac{1}{6} - 2 ln (3))

Decimal

0.09776 \dots

Solution steps

\int_{2}^{3} \frac{2}{x ^{3} + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{3} \frac{1}{x ^{3} + x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} + x ^{2}} : - \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x + 1}

= 2 \cdot \int_{2}^{3} - \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x + 1} 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

= 2 (- \int_{2}^{3} \frac{1}{x} d x + \int_{2}^{3} \frac{1}{x ^{2}} d x + \int_{2}^{3} \frac{1}{x + 1} d x)

\int_{2}^{3} \frac{1}{x} d x = ln (3) - ln (2)

\int_{2}^{3} \frac{1}{x ^{2}} d x = \frac{1}{6}

\int_{2}^{3} \frac{1}{x + 1} d x = 2 ln (2) - ln (3)

= 2 (- (ln (3) - ln (2)) + \frac{1}{6} + 2 ln (2) - ln (3))

Expand

- (ln (3) - ln (2)) + \frac{1}{6} + 2 ln (2) - ln (3) : 3 ln (2) + \frac{1}{6} - 2 ln (3)

= 2 (3 ln (2) + \frac{1}{6} - 2 ln (3))

\int_{- 2}^{2} (9 - ∣ x ∣^{3}) d x

\int_{0}^{6} \frac{x}{3} d x

\int_{0}^{2} \frac{x ^{2}}{3} d x

\int_{0}^{2} (6 x - 3) (8 x^{2} + 2) d x

\int_{1}^{3} \frac{5}{x} - \frac{5}{x ^{2}} d x

What is the integral from 2 to 3 of 2/(x^3+x^2) ?
The integral from 2 to 3 of 2/(x^3+x^2) is 2(3ln(2)+1/6-2ln(3))