Answer to Question #176987 in Calculus for Phyroe

Question

Integrals Giving Inverse Trigonometric Functions

∫(xdx)/(25+16x⁴)

Answer 1

$\displaystyle\int\frac{xdx}{25+16x^4}$

Note that if

$d(x^2)=(x^2)'dx=2xdx$

then

$xdx=\dfrac{1}{2}\cdot d(x^2)$

$\displaystyle\int\frac{xdx}{25+16x^4}=$

$\displaystyle\int\frac{\frac{1}{2}\cdot d(x^2)}{25+16x^4}=$

$\displaystyle\frac{1}{2}\cdot \int\frac{d(x^2)}{25+16x^4}=$

$\displaystyle\frac{1}{2}\cdot \int\frac{d(x^2)}{25+16(x^2)^2}=$

$\displaystyle\frac{1}{2}\cdot \int\frac{d(x^2)}{16\cdot(\frac{25}{16}+(x^2)^2)}=$

$\displaystyle\frac{1}{2\cdot16}\cdot \int\frac{d(x^2)}{\frac{25}{16}+(x^2)^2}=$

$\displaystyle\frac{1}{32}\cdot \int\frac{d(x^2)}{\frac{25}{16}+(x^2)^2}=$ use antiderivative $\displaystyle\int\frac{dt}{a^2+t^2}=\frac{1}{a}\arctan\left(\frac{t}{a}\right)+c$

substituting $t=x^2$ and $a^2=\dfrac{25}{16}$

$\displaystyle\frac{1}{32}\cdot\frac{1}{\sqrt{\frac{25}{16}}}\cdot\arctan\left(\frac{x^2}{\sqrt{\frac{25}{16}}}\right)+c=$

$\displaystyle\frac{1}{32}\cdot\frac{1}{\frac{5}{4}}\cdot\arctan\left(\frac{x^2}{\frac{5}{4}}\right)+c=$

$\displaystyle\frac{1}{32}\cdot\frac{4}{5}\cdot\arctan\left(\frac{4x^2}{5}\right)+c=$

$\displaystyle\frac{1}{8}\cdot\frac{1}{5}\cdot\arctan\left(\frac{4x^2}{5}\right)+c=$

$\displaystyle\frac{1}{40}\cdot\arctan\left(\frac{4x^2}{5}\right)+c$