What is the integral of 16t+sec^2(t) ?

The integral of 16t+sec^2(t) is 8t^2+tan(t)+C

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integral of 16t+sec^2(t)

\int 16 t + sec^{2} (t) d t

8 t^{2} + tan (t) + C

Solution steps

\int 16 t + sec^{2} (t) 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 16 t d t + \int sec^{2} (t) d t

\int 16 t d t = 8 t^{2}

\int sec^{2} (t) d t = tan (t)

= 8 t^{2} + tan (t)

Add a constant to the solution

= 8 t^{2} + tan (t) + C

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9 y \frac{d y}{d x} - 4 x = 0

\frac{d}{d x} (\frac{3}{( 8 x + 3 ) ^{2}})

\frac{d}{d n} (sin (n^{3} m))

\int \frac{1}{1 - 2 x} d x

What is the integral of 16t+sec^2(t) ?
The integral of 16t+sec^2(t) is 8t^2+tan(t)+C