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

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

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

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

e + \frac{1}{3}

Decimal Notation

3.05161 \dots

Solution steps

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

Simplify

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

Expand

\frac{4 ln ^{2} ( x ) + x}{x} : \frac{4 ln ^{2} ( x )}{x} + 1

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

\int_{1}^{e} \frac{4 ln ^{2} ( x )}{x} d x = \frac{4}{3}

\int_{1}^{e} 1 d x = e - 1

= \frac{4}{3} + e - 1

Combine the fractions

- 1 + \frac{4}{3} : \frac{1}{3}

= e + \frac{1}{3}

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