What is the integral of 1.1t+6.3t^3 ?

The integral of 1.1t+6.3t^3 is 0.55t^2+1.575t^4+C

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integral of 1.1t+6.3t^3

\int 1.1 t + 6.3 t^{3} d t

0.55 t^{2} + 1.575 t^{4} + C

Solution steps

\int 1.1 t + 6.3 t^{3} 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 1.1 t d t + \int 6.3 t^{3} d t

\int 1.1 t d t = 0.55 t^{2}

\int 6.3 t^{3} d t = 1.575 t^{4}

= 0.55 t^{2} + 1.575 t^{4}

Add a constant to the solution

= 0.55 t^{2} + 1.575 t^{4} + C

s (t) = 48 (6 t^{3} - 5)^{\frac{6}{5}}

e^{x} (5 x^{2} - 10 x + 10)

\frac{d}{d x} (arcsin (sin (17 x)))

\int \frac{x}{x + 60} d x

What is the integral of 1.1t+6.3t^3 ?
The integral of 1.1t+6.3t^3 is 0.55t^2+1.575t^4+C