What is the integral from 0 to 2 of (15)/(x^2+7x+12) ?

The integral from 0 to 2 of (15)/(x^2+7x+12) is 15(ln(5)-ln(3)-ln(3/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (15)/(x^2+7x+12)

\int_{0}^{2} \frac{15}{x ^{2} + 7 x + 12} d x

15 (ln (5) - ln (3) - ln (\frac{3}{2}))

Decimal Notation

1.58040 \dots

Solution steps

\int_{0}^{2} \frac{15}{x ^{2} + 7 x + 12} d x

Take the constant out:

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

= 15 \cdot \int_{0}^{2} \frac{1}{x ^{2} + 7 x + 12} d x

Take the partial fraction of

\frac{1}{x ^{2} + 7 x + 12} : \frac{1}{x + 3} - \frac{1}{x + 4}

= 15 \cdot \int_{0}^{2} \frac{1}{x + 3} - \frac{1}{x + 4} 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

= 15 (\int_{0}^{2} \frac{1}{x + 3} d x - \int_{0}^{2} \frac{1}{x + 4} d x)

\int_{0}^{2} \frac{1}{x + 3} d x = ln (5) - ln (3)

\int_{0}^{2} \frac{1}{x + 4} d x = ln (\frac{3}{2})

= 15 (ln (5) - ln (3) - ln (\frac{3}{2}))

\int_{0}^{\frac{π}{2}} 3 sin^{3} (x) d x

\int_{1}^{4} \frac{ln ( x )}{x ^{6}} d x

\int_{1}^{16} x^{\frac{3}{2}} d x

\int_{0}^{2} [x^{2}] d x

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

What is the integral from 0 to 2 of (15)/(x^2+7x+12) ?
The integral from 0 to 2 of (15)/(x^2+7x+12) is 15(ln(5)-ln(3)-ln(3/2))