Question

Integrate the function : $\int \sqrt{x^2-8x+7}\, dx$ is equal to
(A)$\frac 12 (x-4)\sqrt{x^2-8x+7}+9\log \bigg\arrowvert x-4+ \sqrt{x^2-8x+7}\bigg\arrowvert + C$
(B)$\frac 12 (x+4)\sqrt{x^2-8x+7}+9\log \bigg\arrowvert x+4+ \sqrt{x^2-8x+7}\bigg\arrowvert + C$
(C)$\frac 12 (x-4)\sqrt{x^2-8x+7}-3\sqrt{2}\log \bigg\arrowvert x-4+ \sqrt{x^2-8x+7}\bigg\arrowvert + C$
(D)$\frac 12 (x-4)\sqrt{x^2-8x+7}-\frac{9}{2}\log \bigg\arrowvert x-4+ \sqrt{x^2-8x+7}\bigg\arrowvert + C$

Answer 1

Answer:

option (d) is correct

Step-by-step explanation:

Formula used:

$\int{\sqrt{x^2-a^2}}\:dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}log|x+\sqrt{x^2-a^2}|+c$

Now,

$\int{\sqrt{x^2-8x+7}}\:dx\\\\=\int{\sqrt{x^2-8x+7+9-9}}\:dx\\\\=\int{\sqrt{(x^2-8x+16)-9}}\:dx\\\\=\int{\sqrt{(x-4)^2-3^2}}\:dx\\\\=\frac{(x-4)}{2}\sqrt{(x-4)^2-3^2}-\frac{3^2}{2}.log|(x-4)+\sqrt{(x-4)^2-3^2}|+c\\\\=\frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}.log|(x-4)+\sqrt{x^2-8x+7}|+c$