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

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

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

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

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

Decimal Notation

2.90903 \dots

Solution steps

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

Expand

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

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

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

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

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

\int_{1}^{8} (\frac{x}{8} - \frac{6}{x}) d x

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

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

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

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

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