Question

Evaluate: $\int \frac{5x^4+4x^5}{{(x^5+x+1)}^2}dx$

• A. $x^5+x+1+C$
• B. $\frac{x^5}{x^5+x+1}+C$
• C. $x^{-4}+x^{-5}+C$
• D. $-\frac{x+1}{x^5+x+1}+C$

Answer 1

The correct option is D $-\frac{x+1}{x^5+x+1}+C$

$I=\int \frac{{5x}^4+{4x}^5}{{(x^5+x+1)}^2}dx=\int \frac{{5x}^4+1}{{\left(x^5+x+1\right)}^2}dx+\int \frac{{4x}^5-1}{{(x^5+x+1)}^2}dx$
Let $I=I_1+I_2$
$I_1=\int \frac{{5x}^4+1}{{(x^5+x+1)}^2}dx$
Substitute $x^5+x+1=t\Rightarrow ({5x}^4+1)dx=dt$
$I_1=\int \frac{1}{t^2}dt=-\frac{1}{t}+c=-\frac{1}{x^5+x+1}+c$
$I_2=\int \frac{{4x}^3-\frac{1}{x^2}}{{(x^4+1+\frac{1}{x})}^2}dx$
Substitute $x^4+\frac{1}{x}+1=\upsilon \Rightarrow ({4x}^3-\frac{1}{x^2})dx=dt$
$I_2=\int \frac{1}{{\upsilon }^2}d\upsilon =-\frac{1}{\upsilon }+c=-\frac{1}{x^4+\frac{1}{x}+1}+c=\frac{-x}{x^5+x+1}+c$
$\therefore I=I_1+I_2=\frac{-(x+1)}{x^5+x+1}+c$