What is the solution for (df)/(dt)=4t+(9t)/(sqrt(9-t^2)) ?

The solution for (df)/(dt)=4t+(9t)/(sqrt(9-t^2)) is f(t)=2t^2-9sqrt(9-t^2)+c_{1}

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(df)/(dt)=4t+(9t)/(sqrt(9-t^2))

\frac{df}{d t} = 4 t + \frac{9 t}{9 - t ^{2}}

f (t) = 2 t^{2} - 9 9 - t^{2} + c_{1}

Solution steps

\frac{df}{d t} = 4 t + \frac{9 t}{9 - t ^{2}}

Substitute

\frac{df}{d t}

with

f^{'} (t)

f^{^{'}} (t) = 4 t + \frac{9 t}{9 - t ^{2}}

f^{^{'}} (x) = g (x)

then

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

f (t) = \int 4 t + \frac{9 t}{9 - t ^{2}} d t

\int 4 t + \frac{9 t}{9 - t ^{2}} d t = 2 t^{2} - 9 9 - t^{2} + c_{1}

f (t) = 2 t^{2} - 9 9 - t^{2} + c_{1}

x \cdot y^{2}

x \to 3 lim (- x + 3)

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

\int 2 s d s

\int \frac{29}{29 + e ^{x}} d x

What is the solution for (df)/(dt)=4t+(9t)/(sqrt(9-t^2)) ?
The solution for (df)/(dt)=4t+(9t)/(sqrt(9-t^2)) is f(t)=2t^2-9sqrt(9-t^2)+c_{1}