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

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

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integral of (2y)/1 ^2

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

d^{2} y x^{2} + C

Solution steps

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

Integral of a constant:

\int a d x = a x

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

Simplify

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

Add a constant to the solution

= d^{2} y x^{2} + C

\int - \frac{6 x}{5} d x

d er i v a t i v e (f (x))^{2}

l a pl a ce t r an s f or m 3 sin (3 t)

2 x^{2} y y^{^{'}} + 2 x y^{2} + 1 = 0

\frac{d}{d x} (3 ln (x + 1))