Question

Integrate with respect to $x$:
(i) $\frac{x^3-1}{x^3+x}$ (ii) $\frac{1}{\sqrt{x^2+2x+5}}$

Answer 1

(i) $\int \frac{(x^3-1)dx}{x^3+x}$

$=\int \left[\frac{(x^3-1)-(x+1)}{x^3+x}\right]dx$

$=\int [1-\frac{x+1}{x(x^2+1)}]dx$

$=\int dx-\int \frac{(x+1)}{x(x+1)}dx$

$=\int dx-\int \frac{xdx}{x(x^2+1)}-\int \frac{dx}{x(x^2+1)}$

By using $\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$

$=x-ta{n}^{-1}\left(x\right)-\int \frac{\left[\right(x^2+1)-x^2]}{x(x^2+1)}dx$

$=x-ta{n}^{-1}x-\int \frac{dx}{x}+\int \frac{x}{x^2+1}dx$

$=x-ta{n}^{-1}x-lnx+\frac{1}{2}ln(x^2+1)+C$

$\therefore \int \frac{(x^3-1)}{(x^3+x)}=x-ta{n}^{-1}x-ln\left(\frac{\sqrt{x^2+1}}{x}\right)+C$

(ii) $\int \frac{dx}{\sqrt{x^2+2x+5}}$

$=\int \frac{dx}{\sqrt{(x+1{)}^2+(2{)}^2}}$

$=ln(x+\sqrt{(x+1{)}^2+(2{)}^2})+C$

$=ln(x+\sqrt{x^2+2x+5})+C$

$\therefore \int \frac{dx}{\sqrt{x^2+2x+5}}=ln(x+\sqrt{x^2+2x+5})+C$