What is the integral of (0.24-0.03t)/(sqrt(36+16t-t^2)) ?

The integral of (0.24-0.03t)/(sqrt(36+16t-t^2)) is 0.3sqrt(-0.01t^2+0.16t+0.36)+C

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integral of (0.24-0.03t)/(sqrt(36+16t-t^2))

\int \frac{0.24 - 0.03 t}{36 + 16 t - t ^{2}} d t

0.3 - 0.01 t^{2} + 0.16 t + 0.36 + C

Solution steps

\int \frac{0.24 - 0.03 t}{36 + 16 t - t ^{2}} d t

Expand

\frac{0.24 - 0.03 t}{36 + 16 t - t ^{2}} : \frac{0.24}{36 + 16 t - t ^{2}} - \frac{0.03 t}{36 + 16 t - t ^{2}}

= \int \frac{0.24}{36 + 16 t - t ^{2}} - \frac{0.03 t}{36 + 16 t - 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{0.24}{36 + 16 t - t ^{2}} d t - \int \frac{0.03 t}{36 + 16 t - t ^{2}} d t

\int \frac{0.24}{36 + 16 t - t ^{2}} d t = 0.24 arcsin (0.1 (t - 8))

\int \frac{0.03 t}{36 + 16 t - t ^{2}} d t = 0.06 (- 5 - 0.01 t^{2} + 0.16 t + 0.36 + 4 arcsin (0.1 (t - 8)))

= 0.24 arcsin (0.1 (t - 8)) - 0.06 (- 5 - 0.01 t^{2} + 0.16 t + 0.36 + 4 arcsin (0.1 (t - 8)))

Simplify

0.24 arcsin (0.1 (t - 8)) - 0.06 (- 5 - 0.01 t^{2} + 0.16 t + 0.36 + 4 arcsin (0.1 (t - 8))) : 0.3 - 0.01 t^{2} + 0.16 t + 0.36

= 0.3 - 0.01 t^{2} + 0.16 t + 0.36

Add a constant to the solution

= 0.3 - 0.01 t^{2} + 0.16 t + 0.36 + C

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What is the integral of (0.24-0.03t)/(sqrt(36+16t-t^2)) ?
The integral of (0.24-0.03t)/(sqrt(36+16t-t^2)) is 0.3sqrt(-0.01t^2+0.16t+0.36)+C