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

The integral from 1 to 2 of 1/(xsqrt(4+x^2)) is 1/2 (ln(sqrt(2)-1)-ln(tan(1/2 arctan(1/2))))

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

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

\frac{1}{2} (ln (2 - 1) - ln (tan (\frac{1}{2} arctan (\frac{1}{2}))))

Decimal

0.28113 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{a r c t a n (\frac{1}{2})}^{\frac{π}{4}} \frac{sec ( u )}{2 tan ( u )} d u

= \int_{a r c t a n (\frac{1}{2})}^{\frac{π}{4}} \frac{sec ( u )}{2 tan ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{a r c t a n (\frac{1}{2})}^{\frac{π}{4}} \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

: \int_{t a n (\frac{a r c t a n ( \frac{1}{2} )}{2})}^{2 - 1} \frac{1}{v} d v

= \frac{1}{2} \cdot \int_{t a n (\frac{a r c t a n ( \frac{1}{2} )}{2})}^{2 - 1} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} [ln ∣ v ∣]_{t a n (\frac{a r c t a n ( \frac{1}{2} )}{2})}^{2 - 1}

Compute the boundaries:

ln (2 - 1) - ln (tan (\frac{1}{2} arctan (\frac{1}{2})))

= \frac{1}{2} (ln (2 - 1) - ln (tan (\frac{1}{2} arctan (\frac{1}{2}))))

\int_{0}^{2} (6 x - 6) (4 x^{2} + 5) d x

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\int_{π}^{\frac{3 π}{2}} ∣ sin (x) ∣ d x

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\int_{0}^{π} 2 x sin (2 x) d x

What is the integral from 1 to 2 of 1/(xsqrt(4+x^2)) ?
The integral from 1 to 2 of 1/(xsqrt(4+x^2)) is 1/2 (ln(sqrt(2)-1)-ln(tan(1/2 arctan(1/2))))