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

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

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

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

\frac{arctan ( \frac{1}{2} )}{2}

Solution steps

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

Expand

x (2 + ln^{2} (x)) : 2 x + x ln^{2} (x)

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

Apply u-substitution

: \int_{0}^{1} \frac{1}{2 + u ^{2}} d u

= \int_{0}^{1} \frac{1}{2 + u ^{2}} d u

Apply Integral Substitution

= \int_{0}^{\frac{1}{2}} \frac{1}{2 ( 1 + v ^{2} )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} \frac{1}{1 + v ^{2}} d v

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= \frac{1}{2} [arctan (v)]_{0}^{\frac{1}{2}}

Compute the boundaries:

arctan (\frac{1}{2})

= \frac{1}{2} arctan (\frac{1}{2})

Simplify

= \frac{arctan ( \frac{1}{2} )}{2}

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