Question

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) x4 18x2 4 x5 30x3 20x dx

Answer 1

Your integrand is missing some symbols. My best interpretation is the following integral:

$I=\displaystyle\int\frac{x^4+18x^2+4}{x^5+30x^3+20x}\,\mathrm dx$

Decompose into partial fractions; we're looking for an expansion of the form

$\dfrac{x^4+18x^2+4}{x^5+30x^3+20x}=\dfrac ax+\dfrac{bx^3+cx^2+dx+e}{x^4+30x^2+20}$

Now:

$x^4+18x^2+4=a(x^4+30x^2+20)+(bx^3+cx^2+dx+e)x$

$=(a+b)x^4+cx^3+(30a+d)x^2+ex+20a$

Matching up coefficients tells us that

$\begin{cases}a+b=1\\c=0\\30a+d=18\\e=0\\20a=4\end{cases}\implies a=\dfrac15,b=\dfrac45,d=12$

so that

$I=\displaystyle\frac15\int\frac{\mathrm dx}x+\frac45\int\frac{x^3+15x}{x^4+30x^2+20}\,\mathrm dx$

The integral is trivial:

$\displaystyle\frac15\int\frac{\mathrm dx}x=\frac15\ln|x|+C$

For the second integral, notice that

$\mathrm d(x^4+30x^2+20)=(4x^3+60x)\,\mathrm dx$

Distribute the 4 over the numerator, then substitute $u=x^4+30x^2+20$ and $\mathrm du=(4x^3+60x)\,\mathrm dx$:

$\displaystyle\frac15\int\frac{4x^3+60x}{x^4+30x^2+20}\,\mathrm dx=\frac15\int\frac{\mathrm du}u=\frac15\ln|u|+C=\frac15\ln(x^4+30x^2+20)+C$

So we have

$I=\dfrac15\ln|x|+\dfrac15\ln(x^4+30x^2+20)+C$

and with some simplification,

$I=\boxed{\ln\sqrt[5]{|x^5+30x^3+20x|}+C}$