What is the integral of 4(x^2+1) ?

The integral of 4(x^2+1) is 4((x^3)/3+x)+C

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integral of 4(x^2+1)

\int 4 (x^{2} + 1) d x

4 (\frac{x ^{3}}{3} + x) + C

Solution steps

\int 4 (x^{2} + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x^{2} + 1 d x

Apply the Sum Rule:

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

= 4 (\int x^{2} d x + \int 1 d x)

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

= 4 (\frac{x ^{3}}{3} + x)

Add a constant to the solution

= 4 (\frac{x ^{3}}{3} + x) + C

f (x) = \frac{x ^{2} - 1}{2}

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