What is the integral of 1/(xsqrt(7-x^2)) ?

The integral of 1/(xsqrt(7-x^2)) is -1/(2sqrt(7))(ln|sqrt((7-x^2)/7)+1|-ln|sqrt((7-x^2)/7)-1|)+C

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integral of 1/(xsqrt(7-x^2))

\int \frac{1}{x 7 - x ^{2}} d x

- \frac{1}{2 7} (ln \frac{7 - x ^{2}}{7} + 1 - ln \frac{7 - x ^{2}}{7} - 1) + C

Solution steps

\int \frac{1}{x 7 - x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 7} d u

= \int - \frac{1}{- u ^{2} + 7} 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} + 7} d u

Apply Integral Substitution

= - \int \frac{1}{7 ( - 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}{7} \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}{7} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{7} \frac{ln \frac{7 - x ^{2}}{7} + 1}{2} - \frac{ln \frac{7 - x ^{2}}{7} - 1}{2}

Simplify

- \frac{1}{7} \frac{ln \frac{7 - x ^{2}}{7} + 1}{2} - \frac{ln \frac{7 - x ^{2}}{7} - 1}{2} : - \frac{1}{2 7} (ln \frac{7 - x ^{2}}{7} + 1 - ln \frac{7 - x ^{2}}{7} - 1)

= - \frac{1}{2 7} (ln \frac{7 - x ^{2}}{7} + 1 - ln \frac{7 - x ^{2}}{7} - 1)

Add a constant to the solution

= - \frac{1}{2 7} (ln \frac{7 - x ^{2}}{7} + 1 - ln \frac{7 - x ^{2}}{7} - 1) + C

\int 4 x - 2 d x

\frac{3}{s ^{2} + 2 s + 10}

\frac{1}{s ( s ^{2} + 2 )}

y = 2 C x + C (sin (x) + x cos (x))

\frac{\partial}{\partial x} (\frac{x y}{y - x})

What is the integral of 1/(xsqrt(7-x^2)) ?
The integral of 1/(xsqrt(7-x^2)) is -1/(2sqrt(7))(ln|sqrt((7-x^2)/7)+1|-ln|sqrt((7-x^2)/7)-1|)+C