What is the integral from 1 to 5 of 4((x-1)^3)/(x^2) ?

The integral from 1 to 5 of 4((x-1)^3)/(x^2) is 4(3ln(5)-4/5)

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integral from 1 to 5 of 4((x-1)^3)/(x^2)

\int_{1}^{5} 4 \frac{( x - 1 ) ^{3}}{x ^{2}} d x

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

Decimal

16.11325 \dots

Solution steps

\int_{1}^{5} 4 \cdot \frac{( x - 1 ) ^{3}}{x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int_{1}^{5} \frac{( x - 1 ) ^{3}}{x ^{2}} d x

Expand

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

= 4 \cdot \int_{1}^{5} x - 3 + \frac{3}{x} - \frac{1}{x ^{2}} 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

= 4 (\int_{1}^{5} x d x - \int_{1}^{5} 3 d x + \int_{1}^{5} \frac{3}{x} d x - \int_{1}^{5} \frac{1}{x ^{2}} d x)

\int_{1}^{5} x d x = 12

\int_{1}^{5} 3 d x = 12

\int_{1}^{5} \frac{3}{x} d x = 3 ln (5)

\int_{1}^{5} \frac{1}{x ^{2}} d x = \frac{4}{5}

= 4 (12 - 12 + 3 ln (5) - \frac{4}{5})

Simplify

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

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\int_{x^{2}}^{x^{4}} (2 t - 1)^{3} d t

\int_{0}^{4} 4 x^{2} + x d x

What is the integral from 1 to 5 of 4((x-1)^3)/(x^2) ?
The integral from 1 to 5 of 4((x-1)^3)/(x^2) is 4(3ln(5)-4/5)