What is the integral of (x^2)/(1+4x^2) ?

The integral of (x^2)/(1+4x^2) is 1/8 (-arctan(2x)+2x)+C

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integral of (x^2)/(1+4x^2)

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

\frac{1}{8} (- arctan (2 x) + 2 x) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{u ^{2}}{8 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{u ^{2}}{1 + u ^{2}} d u

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

= \frac{1}{8} \cdot \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

= \frac{1}{8} (- \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

= \frac{1}{8} (- arctan (u) + u)

Substitute back

u = 2 x

= \frac{1}{8} (- arctan (2 x) + 2 x)

Add a constant to the solution

= \frac{1}{8} (- arctan (2 x) + 2 x) + C

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What is the integral of (x^2)/(1+4x^2) ?
The integral of (x^2)/(1+4x^2) is 1/8 (-arctan(2x)+2x)+C