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

The integral from 4 to 5 of (4x^2)/(x-3) is 4(9ln(2)+15/2)

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 (4x^2)/(x-3)

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

4 (9 ln (2) + \frac{15}{2})

Decimal

54.95329 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{2} \frac{( u + 3 ) ^{2}}{u} d u

= 4 \cdot \int_{1}^{2} \frac{( u + 3 ) ^{2}}{u} d u

Expand

\frac{( u + 3 ) ^{2}}{u} : u + 6 + \frac{9}{u}

= 4 \cdot \int_{1}^{2} u + 6 + \frac{9}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int_{1}^{2} u d u + \int_{1}^{2} 6 d u + \int_{1}^{2} \frac{9}{u} d u)

\int_{1}^{2} u d u = \frac{3}{2}

\int_{1}^{2} 6 d u = 6

\int_{1}^{2} \frac{9}{u} d u = 9 ln (2)

= 4 (\frac{3}{2} + 6 + 9 ln (2))

Combine the fractions

6 + \frac{3}{2} : \frac{15}{2}

= 4 (9 ln (2) + \frac{15}{2})

\int_{0}^{4} x (x^{2} + 1)^{2} d x

\int_{0}^{π} (cos (2 x)) d x

\int_{0}^{z} z^{2} - y^{2} d y

\int_{0}^{2} \frac{4 x ^{3}}{5} d x

\int_{1}^{2} x e^{6 x^{2}} d x

What is the integral from 4 to 5 of (4x^2)/(x-3) ?
The integral from 4 to 5 of (4x^2)/(x-3) is 4(9ln(2)+15/2)