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

The integral from 1 to 2 of ((x+1)^2)/x is ln(2)+7/2

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

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

ln (2) + \frac{7}{2}

Decimal Notation

4.19314 \dots

Solution steps

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

Expand

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

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

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

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

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

= \frac{3}{2} + 2 + ln (2)

Combine the fractions

2 + \frac{3}{2} : \frac{7}{2}

= ln (2) + \frac{7}{2}

\frac{d}{d x} (x (ln (x))^{2})

\int \frac{ln ( x ) - 1}{( ln ( x ) ) ^{2}} d x

h \to 0 lim (\frac{( x + h ) ^{- 5} - x ^{- 5}}{h})

\int \frac{x}{x + 1} d x

t^{2} y^{2} + e^{\frac{1}{t}} + (2 y t^{3} + t^{2} y^{3}) y^{^{'}} = 0

What is the integral from 1 to 2 of ((x+1)^2)/x ?
The integral from 1 to 2 of ((x+1)^2)/x is ln(2)+7/2