What is the integral of (x^2)/(x^2+x^4) ?

The integral of (x^2)/(x^2+x^4) is arctan(x)+C

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integral of (x^2)/(x^2+x^4)

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

arctan (x) + C

Solution steps

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

Simplify

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

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

Use the common integral:

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

= arctan (x)

Add a constant to the solution

= arctan (x) + C

\int_{0}^{1} 5 cos (\frac{π t}{2}) d t

y^{^{''}} - 4 y^{^{'}} = 16 t

\int \frac{1}{( a + b x ) ^{\frac{1}{3}}} d x

x^{2} y d x + x^{3} d y = 0

x \to - 2 lim (- 2 x^{3} - 3 x^{2} + 8)

What is the integral of (x^2)/(x^2+x^4) ?
The integral of (x^2)/(x^2+x^4) is arctan(x)+C