What is the integral of (300)/(x^2sqrt(100-x^2)) ?

The integral of (300)/(x^2sqrt(100-x^2)) is -(3sqrt(-x^2+100))/x+C

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integral of (300)/(x^2sqrt(100-x^2))

\int \frac{300}{x ^{2} 100 - x ^{2}} d x

- \frac{3 - x ^{2} + 100}{x} + C

Solution steps

\int \frac{300}{x ^{2} 100 - x ^{2}} d x

Take the constant out:

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

= 300 \cdot \int \frac{1}{x ^{2} 100 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{100 cos ( u ) sin ( u )} d u

= 300 \cdot \int \frac{cot ( u )}{100 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 300 \cdot \frac{1}{100} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= 300 \cdot \frac{1}{100} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= 300 \cdot \frac{1}{100} (- cot (u))

Substitute back

u = arcsin (\frac{1}{10} x)

= 300 \cdot \frac{1}{100} (- cot (arcsin (\frac{1}{10} x)))

Simplify

300 \cdot \frac{1}{100} (- cot (arcsin (\frac{1}{10} x))) : - \frac{3 - x ^{2} + 100}{x}

= - \frac{3 - x ^{2} + 100}{x}

Add a constant to the solution

= - \frac{3 - x ^{2} + 100}{x} + C

\frac{\partial}{\partial x} (cos^{5} (x^{6} y^{7}))

y = 10 (3 x + 1) (1 - 5 x)

\int arccot (3 y) d y

\int x^{3} x d x

What is the integral of (300)/(x^2sqrt(100-x^2)) ?
The integral of (300)/(x^2sqrt(100-x^2)) is -(3sqrt(-x^2+100))/x+C