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

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

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

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

\frac{π}{3}

Decimal

1.04719 \dots

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{3}} \frac{1}{2} d u

= 2 \cdot \int_{0}^{\frac{π}{3}} \frac{1}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} [1 \cdot u]_{0}^{\frac{π}{3}}

Simplify

2 \cdot \frac{1}{2} [1 \cdot u]_{0}^{\frac{π}{3}} : [u]_{0}^{\frac{π}{3}}

= [u]_{0}^{\frac{π}{3}}

Compute the boundaries:

\frac{π}{3}

= \frac{π}{3}

x \to - 1 + lim (\frac{2}{x + 1})

f (x) = 5 cos (x), at x = \frac{π}{6}

\int \frac{sin ( 5 x )}{cos ^{2} ( 5 x )} d x

x \to + 10 lim (\frac{8}{( x - 10 ) ^{2}})

f (x) = 4 x^{3} - \frac{2}{x ^{3}}

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