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

The integral of 1/1+x^2(x^2)/2 is x+(x^5)/(10)+C

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

integral

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

x + \frac{x ^{5}}{10} + C

Solution steps

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

Simplify

\frac{1}{1} + x^{2} \frac{x ^{2}}{2} : 1 + \frac{x ^{4}}{2}

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

\int 1 d x = x

\int \frac{x ^{4}}{2} d x = \frac{x ^{5}}{10}

= x + \frac{x ^{5}}{10}

Add a constant to the solution

= x + \frac{x ^{5}}{10} + C

\frac{- y}{( 1 + y ^{2} )}

\frac{1}{x} (arctan (\frac{1}{x}))

- 4 e^{3} x

tan (\frac{x}{2})

(2 x - x^{2} + 7) e^{2} x

What is the integral of 1/1+x^2(x^2)/2 ?
The integral of 1/1+x^2(x^2)/2 is x+(x^5)/(10)+C