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

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

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integral of (1/x+2y^2x)

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

ln ∣ x ∣ + y^{2} x^{2} + C

Solution steps

\int \frac{1}{x} + 2 y^{2} x 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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + y^{2} x^{2}

Add a constant to the solution

= ln ∣ x ∣ + y^{2} x^{2} + C

y = 9 - x^{2}

\int 18 x^{2} d x

h \to 0 lim (8)

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

\int csc (2 x) cot (2 x) d x

What is the integral of (1/x+2y^2x) ?
The integral of (1/x+2y^2x) is ln|x|+y^2x^2+C