What is the integral from 0 to 9/10 of 1/((100x^2+1)^{3/2)} ?

The integral from 0 to 9/10 of 1/((100x^2+1)^{3/2)} is 9/(10sqrt(82))

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integral from 0 to 9/10 of 1/((100x^2+1)^{3/2)}

\int_{0}^{\frac{9}{10}} \frac{1}{( 100 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{9}{10 82}

Decimal

0.09938 \dots

Solution steps

\int_{0}^{\frac{9}{10}} \frac{1}{( 100 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (9)} \frac{1}{10 sec ( u )} d u

= \int_{0}^{a r c t a n (9)} \frac{1}{10 sec ( u )} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int_{0}^{a r c t a n (9)} \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{10} \cdot \int_{0}^{a r c t a n (9)} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{10} [sin (u)]_{0}^{a r c t a n (9)}

Compute the boundaries:

\frac{9}{82}

= \frac{1}{10} \cdot \frac{9}{82}

Simplify

= \frac{9}{10 82}

{e^{3 t} sin (6 t) - t^{3} + e^{t}}

12 x^{3} - 3 y^{2} x^{4} + 1 \cdot y^{^{'}} = 0

\int \frac{1}{( x ^{2} - 49 ) ^{2}} d x

f (x) = 2 x + 2

\frac{d y}{d x} = 0.0016 y (2400 - y)

What is the integral from 0 to 9/10 of 1/((100x^2+1)^{3/2)} ?
The integral from 0 to 9/10 of 1/((100x^2+1)^{3/2)} is 9/(10sqrt(82))