If \( \int\displaystyle\frac{(\sqrt{x})^5}{(\sqrt{x})^7+x^6}\,dx=a\log\left|\displaystyle\frac{x^k}{1+x^k}\right|+\mathrm{C} \); where C is an arbitrary constant, then the value of \( a.k \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
Since \( (\sqrt{x})^5=x^{5/2} \) and \( (\sqrt{x})^7=x^{7/2} \), dividing the numerator and the denominator by \( x^{5/2} \) gives
\( \displaystyle \frac{x^{5/2}}{x^{7/2}+x^6}=\frac{1}{x+x^{7/2}} \)
\( \displaystyle =\frac{1}{x\left(1+x^{5/2}\right)} \)
Multiplying the top and bottom by \( x^{3/2} \) brings in the derivative of \( x^{5/2} \):
\( \displaystyle \int\frac{dx}{x\left(1+x^{5/2}\right)}=\int\frac{x^{3/2}\,dx}{x^{5/2}\left(1+x^{5/2}\right)} \)
Put \( t=x^{5/2} \), \( dt=\frac{5}{2}x^{3/2}\,dx \), so \( x^{3/2}\,dx=\frac{2}{5}\,dt \):
\( \displaystyle =\frac{2}{5}\int\frac{dt}{t(1+t)} \)
\( \displaystyle =\frac{2}{5}\int\left(\frac{1}{t}-\frac{1}{1+t}\right)dt \)
\( \displaystyle =\frac{2}{5}\log\left|\frac{t}{1+t}\right|+\mathrm{C} \)
\( \displaystyle =\frac{2}{5}\log\left|\frac{x^{5/2}}{1+x^{5/2}}\right|+\mathrm{C} \)
Comparing, \( a=\frac{2}{5} \) and \( k=\frac{5}{2} \):
\( \displaystyle a\cdot k=\frac{2}{5}\times\frac{5}{2}=1 \)