Final answer to the problem
$\frac{1}{8}\ln\left|1+4x^2\right|+\frac{-\sqrt{\arctan\left(2x\right)^{3}}}{3}+C_0$
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1
Expand the fraction $\frac{x-\sqrt{\arctan\left(2x\right)}}{1+4x^2}$ into $2$ simpler fractions with common denominator $1+4x^2$
$\int\left(\frac{x}{1+4x^2}+\frac{-\sqrt{\arctan\left(2x\right)}}{1+4x^2}\right)dx$
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$\int\left(\frac{x}{1+4x^2}+\frac{-\sqrt{\arctan\left(2x\right)}}{1+4x^2}\right)dx$
Learn how to solve problems step by step online. Integrate int((x-arctan(2x)^(1/2))/(1+4x^2))dx. Expand the fraction \frac{x-\sqrt{\arctan\left(2x\right)}}{1+4x^2} into 2 simpler fractions with common denominator 1+4x^2. Simplify the expression. The integral \int\frac{x}{1+4x^2}dx results in: \frac{1}{8}\ln\left(1+4x^2\right). The integral -\int\frac{\sqrt{\arctan\left(2x\right)}}{1+4x^2}dx results in: \frac{-\sqrt{\arctan\left(2x\right)^{3}}}{3}.
Final answer to the problem
$\frac{1}{8}\ln\left|1+4x^2\right|+\frac{-\sqrt{\arctan\left(2x\right)^{3}}}{3}+C_0$
