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

The integral from 1 to 4 of ((3x^2+2x+1))/x is 2ln(2)+57/2

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

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

2 ln (2) + \frac{57}{2}

Decimal

29.88629 \dots

Solution steps

\int_{1}^{4} \frac{3 x ^{2} + 2 x + 1}{x} d x

Expand

\frac{3 x ^{2} + 2 x + 1}{x} : 3 x + 2 + \frac{1}{x}

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

\int_{1}^{4} 3 x d x = \frac{45}{2}

\int_{1}^{4} 2 d x = 6

\int_{1}^{4} \frac{1}{x} d x = 2 ln (2)

= \frac{45}{2} + 6 + 2 ln (2)

Combine the fractions

6 + \frac{45}{2} : \frac{57}{2}

= 2 ln (2) + \frac{57}{2}

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What is the integral from 1 to 4 of ((3x^2+2x+1))/x ?
The integral from 1 to 4 of ((3x^2+2x+1))/x is 2ln(2)+57/2