What is the integral from 4 to 5 of (x+4)/(x^2-3x) ?

The integral from 4 to 5 of (x+4)/(x^2-3x) is (15ln(2)-4ln(5))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 5 of (x+4)/(x^2-3x)

\int_{4}^{5} \frac{x + 4}{x ^{2} - 3 x} d x

\frac{15 ln ( 2 ) - 4 ln ( 5 )}{3}

Decimal

1.31981 \dots

Solution steps

\int_{4}^{5} \frac{x + 4}{x ^{2} - 3 x} d x

Apply Integration By Parts

: [(x + 4) (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣) - \int \frac{1}{3} (- ln ∣ x ∣ + ln ∣ x - 3 ∣) d x]_{4}^{5}

= [(x + 4) (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣) - \int \frac{1}{3} (- ln ∣ x ∣ + ln ∣ x - 3 ∣) d x]_{4}^{5}

\int \frac{1}{3} (- ln ∣ x ∣ + ln ∣ x - 3 ∣) d x = \frac{1}{3} (- x ln ∣ x ∣ + x ln ∣ x - 3 ∣ - 3 ln ∣ x - 3 ∣ + 3)

= [(x + 4) (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣) - \frac{1}{3} (- x ln ∣ x ∣ + x ln ∣ x - 3 ∣ - 3 ln ∣ x - 3 ∣ + 3)]_{4}^{5}

Simplify

[(x + 4) (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣) - \frac{1}{3} (- x ln ∣ x ∣ + x ln ∣ x - 3 ∣ - 3 ln ∣ x - 3 ∣ + 3)]_{4}^{5} : [\frac{1}{3} (- 4 ln ∣ x ∣ + 7 ln ∣ x - 3 ∣) - 1]_{4}^{5}

= [\frac{1}{3} (- 4 ln ∣ x ∣ + 7 ln ∣ x - 3 ∣) - 1]_{4}^{5}

Compute the boundaries:

\frac{15 ln ( 2 ) - 4 ln ( 5 )}{3}

= \frac{15 ln ( 2 ) - 4 ln ( 5 )}{3}

\int_{0}^{π} 2 sin (x) d x

\int_{1}^{4} (x - x) d x

\int_{1}^{3} (x + 4)^{4} - (x + 1)^{3} d x

\int_{1}^{e^{\frac{π}{2}}} sin (ln (x)) d x

\int_{0}^{5} 9 d x

What is the integral from 4 to 5 of (x+4)/(x^2-3x) ?
The integral from 4 to 5 of (x+4)/(x^2-3x) is (15ln(2)-4ln(5))/3