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

The integral from 2sqrt(2) to 4 of 1/(t^3sqrt(t^2-4)) is (pi+3(sqrt(3)-2))/(192)

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integral from 2sqrt(2) to 4 of 1/(t^3sqrt(t^2-4))

\int_{22}^{4} \frac{1}{t ^{3} t ^{2} - 4} d t

\frac{π + 3 ( 3 - 2 )}{192}

Decimal Notation

0.01217 \dots

Solution steps

\int_{22}^{4} \frac{1}{t ^{3} t ^{2} - 4} d t

Apply Trigonometric Substitution

: \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{8 sec ^{2} ( u )} d u

= \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{8 sec ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{sec ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{2} \cdot \int_{\frac{π}{4}}^{\frac{π}{3}} 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} \cdot \frac{1}{2} (\int_{\frac{π}{4}}^{\frac{π}{3}} 1 d u + \int_{\frac{π}{4}}^{\frac{π}{3}} cos (2 u) d u)

\int_{\frac{π}{4}}^{\frac{π}{3}} 1 d u = \frac{π}{12}

\int_{\frac{π}{4}}^{\frac{π}{3}} cos (2 u) d u = \frac{3 - 2}{4}

= \frac{1}{8} \cdot \frac{1}{2} (\frac{π}{12} + \frac{3 - 2}{4})

Simplify

\frac{1}{8} \cdot \frac{1}{2} (\frac{π}{12} + \frac{3 - 2}{4}) : \frac{π + 3 ( 3 - 2 )}{192}

= \frac{π + 3 ( 3 - 2 )}{192}

y^{^{'}} + y cos (x) = 0

x + 1

\int_{1}^{2} \frac{x ^{2} + 8}{3 x - x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 1})

\frac{7 + 2 x}{5 - 8 x}

What is the integral from 2sqrt(2) to 4 of 1/(t^3sqrt(t^2-4)) ?
The integral from 2sqrt(2) to 4 of 1/(t^3sqrt(t^2-4)) is (pi+3(sqrt(3)-2))/(192)