What is the integral from 1 to 6x of (31+t^2) ?

The integral from 1 to 6x of (31+t^2) is 186x-31+72x^3-1/3

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integral from 1 to 6x of (31+t^2)

\int_{1}^{6 x} (31 + t^{2}) d t

186 x - 31 + 72 x^{3} - \frac{1}{3}

Solution steps

\int_{1}^{6 x} 31 + t^{2} 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}^{6 x} 31 d t + \int_{1}^{6 x} t^{2} d t

\int_{1}^{6 x} 31 d t = 186 x - 31

\int_{1}^{6 x} t^{2} d t = 72 x^{3} - \frac{1}{3}

= 186 x - 31 + 72 x^{3} - \frac{1}{3}

\int_{0}^{2} 30 x d x

\int_{0}^{1} \frac{7 x}{( 2 x + 1 ) ^{3}} d x

\int_{0}^{t} (t - x)^{3} e^{x} d x

\int_{5}^{\infty} e^{- 4 x} d x

\int_{1}^{e} \frac{y ^{2} + 2}{y} d y

What is the integral from 1 to 6x of (31+t^2) ?
The integral from 1 to 6x of (31+t^2) is 186x-31+72x^3-1/3