What is the integral of (3t-2)/(sqrt(9-t^2)) ?

The integral of (3t-2)/(sqrt(9-t^2)) is -3sqrt(9-t^2)-2arcsin(1/3 t)+C

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integral of (3t-2)/(sqrt(9-t^2))

\int \frac{3 t - 2}{9 - t ^{2}} d t

- 3 9 - t^{2} - 2 arcsin (\frac{1}{3} t) + C

Solution steps

\int \frac{3 t - 2}{9 - t ^{2}} d t

Expand

\frac{3 t - 2}{9 - t ^{2}} : \frac{3 t}{9 - t ^{2}} - \frac{2}{9 - t ^{2}}

= \int \frac{3 t}{9 - t ^{2}} - \frac{2}{9 - t ^{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 t}{9 - t ^{2}} d t - \int \frac{2}{9 - t ^{2}} d t

\int \frac{3 t}{9 - t ^{2}} d t = - 3 9 - t^{2}

\int \frac{2}{9 - t ^{2}} d t = 2 arcsin (\frac{1}{3} t)

= - 3 9 - t^{2} - 2 arcsin (\frac{1}{3} t)

Add a constant to the solution

= - 3 9 - t^{2} - 2 arcsin (\frac{1}{3} t) + C

What is the integral of (3t-2)/(sqrt(9-t^2)) ?
The integral of (3t-2)/(sqrt(9-t^2)) is -3sqrt(9-t^2)-2arcsin(1/3 t)+C