What is the integral of 1/(xsqrt(16-x^2)) ?

The integral of 1/(xsqrt(16-x^2)) is -1/8 (ln|(sqrt(16-x^2))/4+1|-ln|(sqrt(16-x^2))/4-1|)+C

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integral of 1/(xsqrt(16-x^2))

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

- \frac{1}{8} (ln \frac{16 - x ^{2}}{4} + 1 - ln \frac{16 - x ^{2}}{4} - 1) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 16} d u

= \int - \frac{1}{- u ^{2} + 16} d u

Take the constant out:

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

= - \int \frac{1}{- u ^{2} + 16} d u

Apply Integral Substitution

= - \int \frac{1}{4 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{4} \frac{ln \frac{16 - x ^{2}}{4} + 1}{2} - \frac{ln \frac{16 - x ^{2}}{4} - 1}{2}

Simplify

- \frac{1}{4} \frac{ln \frac{16 - x ^{2}}{4} + 1}{2} - \frac{ln \frac{16 - x ^{2}}{4} - 1}{2} : - \frac{1}{8} (ln \frac{16 - x ^{2}}{4} + 1 - ln \frac{16 - x ^{2}}{4} - 1)

= - \frac{1}{8} (ln \frac{16 - x ^{2}}{4} + 1 - ln \frac{16 - x ^{2}}{4} - 1)

Add a constant to the solution

= - \frac{1}{8} (ln \frac{16 - x ^{2}}{4} + 1 - ln \frac{16 - x ^{2}}{4} - 1) + C

f^{9}

\int \frac{7 x ^{3}}{4 + x ^{2}} d x

y = - 2 x^{3}, (1, - 2)

\int (3 x^{2}) d x

\frac{\partial}{\partial x} (sin^{2} (3 x y^{2} - y))

What is the integral of 1/(xsqrt(16-x^2)) ?
The integral of 1/(xsqrt(16-x^2)) is -1/8 (ln|(sqrt(16-x^2))/4+1|-ln|(sqrt(16-x^2))/4-1|)+C