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

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

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

\int \frac{x ^{2}}{2 x ^{2} + 2} d x

\frac{1}{2} (- arctan (x) + x) + C

Solution steps

\int \frac{x ^{2}}{2 x ^{2} + 2} d x

\frac{x ^{2}}{2 x ^{2} + 2} = \frac{1}{2} \frac{x ^{2}}{x ^{2} + 1}

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

Take the constant out:

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

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

\frac{x ^{2}}{x ^{2} + 1} = - \frac{1}{x ^{2} + 1} + 1

= \frac{1}{2} \cdot \int - \frac{1}{x ^{2} + 1} + 1 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

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

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

\int 1 d x = x

= \frac{1}{2} (- arctan (x) + x)

Add a constant to the solution

= \frac{1}{2} (- arctan (x) + x) + C

\int 9 arcsin (x) d x

f (x) = 3 e^{3 x}

\frac{d}{d x} (x^{(\frac{4}{5})} (x - 4)^{2})

\frac{d y}{d x} = \frac{x}{4 + x ^{2}}

\frac{d}{d x} (2^{x} ln (x))

What is the integral of (x^2)/(2x^2+2) ?
The integral of (x^2)/(2x^2+2) is 1/2 (-arctan(x)+x)+C