What is the integral of 6t+tsin(t^2+1) ?

The integral of 6t+tsin(t^2+1) is 3t^2-1/2 cos(t^2+1)+C

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integral of 6t+tsin(t^2+1)

\int 6 t + t sin (t^{2} + 1) d t

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

Solution steps

\int 6 t + t sin (t^{2} + 1) 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 6 t d t + \int t sin (t^{2} + 1) d t

\int 6 t d t = 3 t^{2}

\int t sin (t^{2} + 1) d t = - \frac{1}{2} cos (t^{2} + 1)

= 3 t^{2} - \frac{1}{2} cos (t^{2} + 1)

Add a constant to the solution

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

What is the integral of 6t+tsin(t^2+1) ?
The integral of 6t+tsin(t^2+1) is 3t^2-1/2 cos(t^2+1)+C