What is the solution for f^'(t)= 6/(t^6)-3/(t^3)+2/t ?

The solution for f^'(t)= 6/(t^6)-3/(t^3)+2/t is f(t)=-6/(5t^5)+3/(2t^2)+2ln(t)+c_{1}

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f^'(t)= 6/(t^6)-3/(t^3)+2/t

f^{^{'}} (t) = \frac{6}{t ^{6}} - \frac{3}{t ^{3}} + \frac{2}{t}

f (t) = - \frac{6}{5 t ^{5}} + \frac{3}{2 t ^{2}} + 2 ln (t) + c_{1}

Solution steps

f^{'} (t) = \frac{6}{t ^{6}} - \frac{3}{t ^{3}} + \frac{2}{t}

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

then

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

f (t) = \int \frac{6}{t ^{6}} - \frac{3}{t ^{3}} + \frac{2}{t} d t

\int \frac{6}{t ^{6}} - \frac{3}{t ^{3}} + \frac{2}{t} d t = - \frac{6}{5 t ^{5}} + \frac{3}{2 t ^{2}} + 2 ln (t) + c_{1}

f (t) = - \frac{6}{5 t ^{5}} + \frac{3}{2 t ^{2}} + 2 ln (t) + c_{1}

\frac{d}{d x} (3 sin (x))

f (y) = sin (y)

x \to 1 lim (e^{2 x})

f (x) = - 4 x^{2} - 4, g (x) = 2 x^{2} - 9

\frac{d}{d x} (- 3 ln (cos (\frac{x}{3})))

What is the solution for f^'(t)= 6/(t^6)-3/(t^3)+2/t ?
The solution for f^'(t)= 6/(t^6)-3/(t^3)+2/t is f(t)=-6/(5t^5)+3/(2t^2)+2ln(t)+c_{1}