What is the integral of 4/(x^2sqrt(16-x^2)) ?

The integral of 4/(x^2sqrt(16-x^2)) is -(sqrt(16-x^2))/(4x)+C

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integral of 4/(x^2sqrt(16-x^2))

\int \frac{4}{x ^{2} 16 - x ^{2}} d x

- \frac{16 - x ^{2}}{4 x} + C

Solution steps

\int \frac{4}{x ^{2} 16 - x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} 16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

= 4 \cdot \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{16} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= 4 \cdot \frac{1}{16} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 4 \cdot \frac{1}{16} (- cot (u))

Substitute back

u = arcsin (\frac{1}{4} x)

= 4 \cdot \frac{1}{16} (- cot (arcsin (\frac{1}{4} x)))

Simplify

4 \cdot \frac{1}{16} (- cot (arcsin (\frac{1}{4} x))) : - \frac{16 - x ^{2}}{4 x}

= - \frac{16 - x ^{2}}{4 x}

Add a constant to the solution

= - \frac{16 - x ^{2}}{4 x} + C

(x + 2)^{2} + (y - 3)^{2} = 37, (4, 4)

\int \frac{x}{1} d x

x = 5 y^{2}, x = 1 + 4 y^{2}

3 x^{- \frac{3}{2}} + 6 x^{- \frac{1}{2}} + x^{3} - 4

y = e^{3 x}

What is the integral of 4/(x^2sqrt(16-x^2)) ?
The integral of 4/(x^2sqrt(16-x^2)) is -(sqrt(16-x^2))/(4x)+C