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

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

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integral of (y+1)x^2

\int (y + 1) x^{2}

x^{2} (y + \frac{y ^{2}}{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply the Sum Rule:

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

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

\int 1 d y = y

\int y d y = \frac{y ^{2}}{2}

= x^{2} (y + \frac{y ^{2}}{2})

Add a constant to the solution

= x^{2} (y + \frac{y ^{2}}{2}) + C

sin (π t)

y = 8 x, y = 8 x^{3}

\frac{d}{d x} (sin (y) + x y - x^{3})

\frac{d}{d x} ((x + 1)^{l n (x)})

y^{^{''}} + 2 y^{^{'}} = 2 x + 7 - e^{- 2 x}