Question

Integration of 1/ (x 4 + a 4 ) dx

Answer 1

Integration:  I= f(x) dx = dx / (x⁴ + a⁴)
Let   x = a * t    => I= f(t) dt = (1/a³) dt / (1 + t⁴)
We use partial fractions to simplify the expression.

$x=a t,\ \ dx=a\ dt,\ \ I= \int {\frac{dx}{x^4+a^4}}=\frac{1}{a^3}\int {\frac{1}{t^4+1}}\\\\\frac{1}{t^4+1}=\frac{1}{(t^2+1)^2-(\sqrt2\ t)^2}=\frac{At+B}{t^2+\sqrt2t+1}+\frac{-At+D}{t^2-\sqrt2t+1}\\\\(-\sqrt2A+B-\sqrt2A+D)t^2+(A-\sqrt2B-A+\sqrt2D)t+B+D=1\\\\B=D=1/2,\ \ A=-\frac{1}{2\sqrt2}$

$2\sqrt2a^3*I=\int{\frac{(t+\sqrt2)dt}{(t-\frac{1}{\sqrt2})^2+(\frac{1}{\sqrt2})^2}}-\int{\frac{(t-\sqrt2)dt}{(t+\frac{1}{\sqrt2})^2+(\frac{1}{\sqrt2})^2}}\\\\u=\frac{t-\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=\sqrt2t-1,\ \ t=\frac{u+1}{\sqrt2},\ dt=\frac{du}{\sqrt2}\\\\z=\sqrt2t+1,\ t=\frac{z-1}{\sqrt2},\ \ dt=\frac{dz}{\sqrt2}\\\\2\sqrt2a^3*I=\int{\frac{u+3}{u^2+1}}du-\int{\frac{z-3}{z^2+1}}dz\\\\=\frac{1}{2}Ln[\frac{u^2+1}{z^2+1}]+3[Tan^{-1}u+Tan^{-1}z]+K$

$I=\frac{1}{4\sqrt2a^3}Ln[\frac{x^2-\sqrt2ax+a^2}{x^2+\sqrt2ax+a^2}]+\frac{3}{2\sqrt2a^3}[Tan^{-1}(\frac{\sqrt2x}{a}-1)+Tan^{-1}(\frac{\sqrt2x}{a}+1)]\\\\=\frac{1}{4\sqrt2a^3}Ln[\frac{x^2-\sqrt2ax+a^2}{x^2+\sqrt2ax+a^2}]+\frac{3}{2\sqrt2a^3}Tan^{-1}(\frac{\sqrt2ax}{a^2-x^2})+K$

This is the answer.

Answer 2

Answer:

See the attachment.

Hope it's helpful!