What is the integral of 1/(tsqrt(25-t^2)) ?

The integral of 1/(tsqrt(25-t^2)) is -1/10 (ln|(sqrt(25-t^2))/5+1|-ln|(sqrt(25-t^2))/5-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(tsqrt(25-t^2))

\int \frac{1}{t 25 - t ^{2}} d t

- \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1) + C

Solution steps

\int \frac{1}{t 25 - t ^{2}} d t

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 25} d u

= \int - \frac{1}{- u ^{2} + 25} 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} + 25} d u

Apply Integral Substitution

= - \int \frac{1}{5 ( - 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}{5} \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}{5} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{5} \frac{ln \frac{25 - t ^{2}}{5} + 1}{2} - \frac{ln \frac{25 - t ^{2}}{5} - 1}{2}

Simplify

- \frac{1}{5} \frac{ln \frac{25 - t ^{2}}{5} + 1}{2} - \frac{ln \frac{25 - t ^{2}}{5} - 1}{2} : - \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1)

= - \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1)

Add a constant to the solution

= - \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1) + C

\int 20 sin^{3} (x) cos^{2} (x) d x

\int x^{3} ln ∣ 2 x ∣ d x

\int 6 t^{2} + 1 d t

\int \frac{( x + 3 ) ^{3}}{x} d x

\int e^{- s t} (cos (t)) d t

What is the integral of 1/(tsqrt(25-t^2)) ?
The integral of 1/(tsqrt(25-t^2)) is -1/10 (ln|(sqrt(25-t^2))/5+1|-ln|(sqrt(25-t^2))/5-1|)+C