$$\displaystyle \int{dx\over{x(x^4-1)}}$$
Can this integral be calculated using the Partial Fractions method.
Answer
HINT:
We need to use Partial Fraction Decomposition
Method $1:$
As $x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1),$
$$\text{Put }\frac1{x(x^4-1)}=\frac Ax+\frac B{x-1}+\frac C{x+1}+\frac {Dx+E}{x^2+1}$$
Method $2:$
$$I=\int \frac1{x(x^4-1)}dx=\int \frac{xdx}{x^2(x^4-1)} $$
Putting $x^2=y,2xdx=dy,$
$$I=\frac12\int \frac{dy}{y(y^2-1)}$$
$$\text{ Now, put }\frac1{y(y^2-1)}=\frac A y+\frac B{y-1}+\frac C{y+1}$$
Method $3:$
$$I=\int \frac1{x(x^4-1)}dx=\int \frac{x^3dx}{x^4(x^4-1)} $$
Putting $x^4=z,4x^3dx=dz,$
$$I=\frac14\int \frac{dz}{z(z-1)}$$
$$\text{ Now, put }\frac1{z(z-1)}=\frac Az+\frac B{z-1}$$
$$\text{ or by observation, }\frac1{z(z-1)}=\frac{z-(z-1)}{z(z-1)}=\frac1{z-1}-\frac1z$$
Observe that the last method is susceptible to generalization.
$$J=\int\frac{dx}{x(x^n-a)}=\int\frac{x^{n-1}dx}{x^n(x^n-a)}$$
Putting $x^n=u,nx^{n-1}dx=du,$
$$J=\frac1n\int \frac{du}{ u(u-a)}$$
$$\text{ and }\frac1{u(u-a)}=\frac1a\cdot\frac{u-(u-a)}{u(u-a)}=\frac1a\left(\frac1{u-a}-\frac1u\right)$$
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