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

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

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

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

- arctan (u) + u + C

Solution steps

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

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

= \int - \frac{1}{1 + u ^{2}} + 1 d u

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}{1 + u ^{2}} d u + \int 1 d u

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

\int 1 d u = u

= - arctan (u) + u

Add a constant to the solution

= - arctan (u) + u + C

y = 2 x^{2}, y = 8 x^{2}, 5 x + y = 3

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

f (x) = 8 + 5 x^{2} - 2 x^{3}, (2, 12)

\int_{7}^{9} \frac{7}{( x - 8 ) ^{10}} d x

\int \frac{( x ^{2} - x + 42 )}{x ^{3} + 7 x} d x

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