Question

Differentiation question
19th and 20th
please solve​

Answer 1

Answer:

19)

$\displaystyle \frac{x^6-1}{x^2+1}=\frac{(x^2)^3-1}{x^2+1}=\frac{(x^2+1)(x^4-x^2+1)}{x^2+1}=x^4-x^2+1$

So...

$\displaystyle\int\frac{x^6-1}{x^2+1}dx=\int(x^4-x^2+1)\,dx=\tfrac15x^5-\tfrac13x^3+x+C$

20)

Notice that e^(a log x) = ( e^(log x) )^a = x^a.  So the expression to be integrated is...

$\displaystyle\frac{x^6-x^5}{x^5-x^3}=\frac{x^3-x^2}{x^2-1}=\frac{x^2(x-1)}{(x+1)(x-1)}\\ \\=\frac{x^2}{x+1}=\frac{(x^2-1)+1}{x+1}=x-1+\frac1{x+1}$

So...

$\displaystyle\int\left(x-1+\frac1{x+1}\right)dx = \tfrac12x^2-x+\ln|x+1| + C$