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

The integral from 1 to 3 of ((x+1)^2)/x is ln(3)+8

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integral from 1 to 3 of ((x+1)^2)/x

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

ln (3) + 8

Decimal

9.09861 \dots

Solution steps

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

Expand

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

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

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

\int_{1}^{3} x d x = 4

\int_{1}^{3} 2 d x = 4

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

= 4 + 4 + ln (3)

Simplify

= ln (3) + 8

What is the integral from 1 to 3 of ((x+1)^2)/x ?
The integral from 1 to 3 of ((x+1)^2)/x is ln(3)+8