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

The integral from 2 to 3 of 1/(xsqrt(x^4-4)) is 1/4 (arctan((sqrt(77))/2)-pi/3)

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

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

\frac{1}{4} (arctan (\frac{77}{2}) - \frac{π}{3})

Decimal

0.07487 \dots

Solution steps

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

Apply u-substitution

= \int_{12}^{77} \frac{1}{4 ( u + 4 ) u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{12}^{77} \frac{1}{( u + 4 ) u} d u

Apply u-substitution

= \frac{1}{4} \cdot \int_{23}^{77} \frac{2}{v ^{2} + 4} d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int_{23}^{77} \frac{1}{v ^{2} + 4} d v

Apply Integral Substitution

= \frac{1}{4} \cdot 2 \cdot \int_{3}^{\frac{77}{2}} \frac{1}{2 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} \cdot \int_{3}^{\frac{77}{2}} \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} [arctan (w)]_{3}^{\frac{77}{2}}

Simplify

\frac{1}{4} \cdot 2 \cdot \frac{1}{2} [arctan (w)]_{3}^{\frac{77}{2}} : \frac{1}{4} [arctan (w)]_{3}^{\frac{77}{2}}

= \frac{1}{4} [arctan (w)]_{3}^{\frac{77}{2}}

Compute the boundaries:

arctan (\frac{77}{2}) - \frac{π}{3}

= \frac{1}{4} (arctan (\frac{77}{2}) - \frac{π}{3})

What is the integral from 2 to 3 of 1/(xsqrt(x^4-4)) ?
The integral from 2 to 3 of 1/(xsqrt(x^4-4)) is 1/4 (arctan((sqrt(77))/2)-pi/3)