What is the integral of (sqrt(3-4t^2))/(t^2) ?

The integral of (sqrt(3-4t^2))/(t^2) is -(sqrt(3-4t^2))/t-2arcsin(2/(sqrt(3))t)+C

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integral of (sqrt(3-4t^2))/(t^2)

\int \frac{3 - 4 t ^{2}}{t ^{2}} d t

- \frac{3 - 4 t ^{2}}{t} - 2 arcsin (\frac{2}{3} t) + C

Solution steps

\int \frac{3 - 4 t ^{2}}{t ^{2}} d t

Apply Integration By Parts

: - \frac{3 - 4 t ^{2}}{t} - \int \frac{4}{3 - 4 t ^{2}} d t

= - \frac{3 - 4 t ^{2}}{t} - \int \frac{4}{3 - 4 t ^{2}} d t

\int \frac{4}{3 - 4 t ^{2}} d t = 2 arcsin (\frac{2}{3} t)

= - \frac{3 - 4 t ^{2}}{t} - 2 arcsin (\frac{2}{3} t)

Add a constant to the solution

= - \frac{3 - 4 t ^{2}}{t} - 2 arcsin (\frac{2}{3} t) + C

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What is the integral of (sqrt(3-4t^2))/(t^2) ?
The integral of (sqrt(3-4t^2))/(t^2) is -(sqrt(3-4t^2))/t-2arcsin(2/(sqrt(3))t)+C