Question

If $\int \frac{dx}{\sqrt{px+q}-\sqrt{px+r}}=\frac{2}{3k}\left[(px+q{)}^{\frac{3}{2}}+(px+r{)}^{\frac{3}{2}}\right]+C,p\ne 0,q\ne r\ne p$, then $k$ is equal to

• A. $p(q-r)$
• B. $p(r-q)$
• C. $q(r-p)$
• D. $q(p-r)$

Answer 1

Answer: (A)

$\int \frac{\sqrt{px+q}+\sqrt{px+r}}{q-r}dx=\frac{1}{q-r}\int \left((px+q{)}^{\frac{1}{2}}+(px+r{)}^{\frac{1}{2}}\right)dx$
$=\left(\frac{1}{q-r}\right)\cdot \frac{2}{3p}\left((px+q{)}^{\frac{3}{2}}+(px+r{)}^{\frac{3}{2}}\right)$