If \( \int\displaystyle\frac{x^3\,dx}{\sqrt{1+x^2}}=a(1+x^2)^{3/2}+b\sqrt{1+x^2}\ +c \) , where ‘\( c \)’ is an arbitrary constant , then the value of \( a \) and \( b \) are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( a=\displaystyle\frac{1}{3},\ b=-1 \)
Let \( t=1+x^2 \), so \( dt=2x\,dx \) and \( x^2=t-1 \).
\( \displaystyle \int\frac{x^3\,dx}{\sqrt{1+x^2}} \)
\( \displaystyle =\int\frac{x^2\cdot x\,dx}{\sqrt{1+x^2}} \)
\( \displaystyle =\frac{1}{2}\int\frac{t-1}{\sqrt{t}}\,dt \)
\( \displaystyle =\frac{1}{2}\int\left(t^{1/2}-t^{-1/2}\right)dt \)
\( \displaystyle =\frac{1}{2}\left(\frac{2}{3}t^{3/2}-2t^{1/2}\right)+c \)
\( \displaystyle =\frac{1}{3}(1+x^2)^{3/2}-\sqrt{1+x^2}+c \)
Comparing, \( a=\frac{1}{3} \) and b = −1.