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Exercise

$\int arctan\left(\frac{1}{2x}\right)dx$

Step-by-step Solution

1

Apply integration by parts: $u=arctan(\frac{1}{2x})$, $v'=1$

$x\arctan\left(\frac{1}{2x}\right)-\int\frac{\frac{1}{2x}}{1+\left(\frac{1}{2x}\right)^2}dx$
2

Simplify the expression

$x\arctan\left(\frac{1}{2x}\right)-\int\frac{\left(2x\right)^2}{2x\left(1+4x^2\right)}dx$
3

The integral $-\int\frac{\left(2x\right)^2}{2x\left(1+4x^2\right)}dx$ results in: $-2\int\frac{x^2}{x+4x^{3}}dx$

$-2\int\frac{x^2}{x+4x^{3}}dx$
4

Gather the results of all integrals

$x\arctan\left(\frac{1}{2x}\right)-2\int\frac{x^2}{x+4x^{3}}dx$
5

Rewrite the expression $\frac{x^2}{x+4x^{3}}$ inside the integral in factored form

$x\arctan\left(\frac{1}{2x}\right)-2\int\frac{x}{1+4x^2}dx$
6

The integral $-2\int\frac{x}{1+4x^2}dx$ results in: $-\frac{1}{4}\ln\left(1+4x^2\right)$

$-\frac{1}{4}\ln\left(1+4x^2\right)$
7

Gather the results of all integrals

$x\arctan\left(\frac{1}{2x}\right)-\frac{1}{4}\ln\left|1+4x^2\right|$
8

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$x\arctan\left(\frac{1}{2x}\right)-\frac{1}{4}\ln\left|1+4x^2\right|+C_0$

Final answer to the exercise

$x\arctan\left(\frac{1}{2x}\right)-\frac{1}{4}\ln\left|1+4x^2\right|+C_0$

Try other ways to solve this exercise

  • Choose an option
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
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