What is the integral of 1/(usqrt(7-u^2)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(usqrt(7-u^2))

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= - \int \frac{1}{7 ( - w ^{2} + 1 )} d w

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}{- w ^{2} + 1} d w

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f^{^{'}} (x) = \frac{8 x ^{5} - 6 x ^{9}}{x ^{4}}

\frac{d}{d x} (ln (1 + e^{2 x}))

x \to - 1 lim (\frac{x ^{2} - x}{x})

\int_{2}^{\infty} \frac{1}{x ^{2} - 1} d x

\frac{d y}{d x} + y = y^{2}, y (0) = 2

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