Question

If $I=\int \frac{dx}{1+\sqrt{x^2+2x+2}}=\frac{A}{7}\log |x+1+\sqrt{x^2+2x+2}|-\frac{\sqrt{x^2+2x+2}-1}{x+1}+C$
then A is equal to

Answer 1

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

Substitute ${(x+1)}^2+1=t^2$

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

$I=\log |t+\sqrt{t^2-1}|-I_1$

Where $I_1=\int \frac{dt}{(t+1)\sqrt{t^2-1}}$

Put $t+1=\frac{1}{u}$

$I_1=\int \frac{(-\frac{1}{u^2})du}{\left(\frac{1}{u}\right)\sqrt{{(\frac{1}{u}-1)}^2-1}}=-\int \frac{du}{\sqrt{1-2u}}=\sqrt{1-2u}=\sqrt{\frac{t-1}{t+1}}$

Therefore,

$I=\log |t+\sqrt{t^2-1}|-\frac{t-1}{\sqrt{t^2-1}}+c$

$I=\log |x+1+\sqrt{x^2+2x+2}|-\frac{\sqrt{x^2+2x+2}-1}{x+1}+c$

Comparing with, $I=\int \frac{dx}{1+\sqrt{x^2+2x+2}}=\frac{A}{7}\log |x+1+\sqrt{x^2+2x+2}|-\frac{\sqrt{x^2+2x+2}-1}{x+1}+C$

Therefore, $A=7$