Question

Find $\int \frac{x^3+x+1}{x^2-1}dx$

Answer 1

$\int \frac{x^{3}x+1}{x^2-1}dx=\int (\frac{x^3}{x^2-1}+\frac{x+1}{x^2-1})dx$

$=\int \frac{x^{3}dx}{x^2-1}+\int \frac{(x+1)dx}{(x+1)(x-1)}$

$=\int \frac{x^{3}dx}{x^2-1}+\int \frac{1dx}{x-1}$

Now $\int \frac{x^{3}dx}{x^2-1}$

Put $x^2-1$

differentiating wrt.x

2x .dx = dt $\therefore xdx=\frac{dt}{2}$

$=\int \frac{x^2.xdx}{x^2-1}=\int \frac{t.dt}{2(t-1)}=\frac{1}{2}\int \frac{t}{t-1}$

$=\frac{1}{2}[\int \frac{t-1}{t-1}dt+\int \frac{1}{t-1}dt]$

$=\frac{1}{2}[\int dt+log(t-1\left)\right]$

$=\frac{t}{2}+\frac{1}{2}log(t-1)$

$=\frac{x^2}{2}+\frac{1}{2}log(x^2-1)+c$

$\therefore \int \frac{x^3+x+1}{x^2-1}=\int \frac{x^{3}dx}{x^2-1}+\int \frac{dx}{x-1}$

$=\frac{x^2}{2}+\frac{1}{2}log(x^2-1)+log(x-1)+c$

$=\frac{x^2}{2}+log\sqrt{x^2-1}+log(x-1)+c$

$=\frac{x^2}{2}+log\left(\sqrt{x^2-1}\right(x-1\left)\right)+c$