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

The integral of (y^2+ln(x)) is ln(x)y+(y^3)/3+C

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integral of (y^2+ln(x))

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

ln (x) y + \frac{y ^{3}}{3} + C

Solution steps

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

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

\int ln (x) d y = ln (x) y

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

= ln (x) y + \frac{y ^{3}}{3}

Add a constant to the solution

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

\int_{0}^{5} 7 x - x^{2} - 2 x d x

x \to 0 lim (sec (x))

\int \frac{x ^{4} + 2 x + 1}{x ^{3} - x ^{2} - 2 x} d x

f (x) = e^{e x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 4, y^{^{'}} (0) = 5

What is the integral of (y^2+ln(x)) ?
The integral of (y^2+ln(x)) is ln(x)y+(y^3)/3+C