Question 238 of 416MediumMCQ
If ∫x41−x2dx=A(x)(1−x2)m+C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals: Solutions · 1
Official solution
\[\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} \] dx = A(x) ({\left( {\sqrt {1 - {x^2}} } \right)^m}\] + C
\[\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,\]
Put \[{1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}\]
Case-I \[x \ge 0\]
\[ - {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C\]
\[ \Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}\]
\[ \Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C\]
\[A(x) = - {1 \over {3{x^3}}}\,\,\] and m=3
\[{(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}\]
Case-II \[x \le 0\]
We get \[{{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C\]
\[A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3\]
\[{(A(x))^m} = {{ - 1} \over {27{x^9}}}\] Answer the question to unlock