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

The integral from x^2 to e^x of x(t^2+1) is x(e^x-x^2+(e^{3x}-x^6)/3)

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

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

x (e^{x} - x^{2} + \frac{e ^{3 x} - x ^{6}}{3})

Solution steps

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

Take the constant out:

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

= x \cdot \int_{x^{2}}^{e^{x}} 1 + 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

= x (\int_{x^{2}}^{e^{x}} 1 d t + \int_{x^{2}}^{e^{x}} t^{2} d t)

\int_{x^{2}}^{e^{x}} 1 d t = e^{x} - x^{2}

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

= x (e^{x} - x^{2} + \frac{e ^{3 x} - x ^{6}}{3})

\int_{1}^{4} (3 x^{2} + 6) d x

\int_{0}^{1.5} (2 x - 3) d x

\int_{0}^{5} x e^{- \frac{x}{2}} d x

\int_{0}^{2} 2 e^{2 z} d z

What is the integral from x^2 to e^x of x(t^2+1) ?
The integral from x^2 to e^x of x(t^2+1) is x(e^x-x^2+(e^{3x}-x^6)/3)