Question

The value of $\int \frac{dt}{t^2+2xt+1}(x^2>1)$is....

• A. $\frac{1}{2\sqrt{(x^2-1)}}\log \frac{t+x-\sqrt{x^2-1}}{t+x+\sqrt{(x^2+1)}}.$
• B. $\frac{1}{2\sqrt{(x^2-1)}}\log \frac{t+x-\sqrt{x^2-1}}{t+x+\sqrt{(x^2-1)}}+c.$
• C. $\frac{1}{2\sqrt{(x^2+1)}}\log \frac{t+x-\sqrt{x^2+1}}{t+x+\sqrt{(x^2+1)}}.$
• D. $\frac{1}{2\sqrt{(x^2+1)}}\log \frac{t+x-\sqrt{x^2+1}}{t+x+\sqrt{(x^2-1)}}+c.$

Answer 1

The correct option is B $\frac{1}{2\sqrt{(x^2-1)}}\log \frac{t+x-\sqrt{x^2-1}}{t+x+\sqrt{(x^2-1)}}+c.$

Let $I=\int \frac{1}{t^2+2tx+1}dt$

$=\int \frac{dt}{(x+t-\sqrt{x^2-1})(x+t+\sqrt{x^2-1})}$ ..... [Factoring denominator]

$=\int \left(\frac{1}{2\sqrt{x^2-1}(x+t-\sqrt{x^2-1})}-\frac{1}{2\sqrt{x^2-1}(x+t+\sqrt{t^2-1})}\right)dt$ .... [Using partial decomposition]

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

$=I_1-I_2$

$I_1=\int \frac{dt}{2\sqrt{x^2-1}(x+t-\sqrt{x^2-1})}$
Put $t+x-\sqrt{x^2-1}=u\Rightarrow dt=du$

$⟹I_1=\frac{1}{2\sqrt{x^2-1}}\int \frac{du}{u}$

$⟹I_1=\frac{1}{2\sqrt{x^2-1}}\log \left(u\right)$

$⟹I_1=\frac{1}{2\sqrt{x^2-1}}\log (x+t-\sqrt{x^2-1})$

Similarly,

$I_2=\int \frac{dt}{2\sqrt{x^2-1}(x+t+\sqrt{x^2-1})}$
Put $t+x+\sqrt{x^2-1}=u\Rightarrow dt=du$

$⟹I_2=\frac{1}{2\sqrt{x^2-1}}\int \frac{du}{u}$

$⟹I_2=\frac{1}{2\sqrt{x^2-1}}\log \left(u\right)$

$⟹I_2=\frac{1}{2\sqrt{x^2-1}}\log (x+t+\sqrt{x^2-1})$

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

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

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