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Question

If $$\displaystyle\int { { x }^{ 13/2 }.{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 1/2 }dx } =P{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 7/2 }+Q{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 5/2 }+R{ \left( 1+{ x }^{ 5/2 } \right)  }^{ 3/2 }+C$$, then $$P,\ Q$$ and $$R$$ are


A
P=435, Q=825, R=415
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B
P=435, Q=825, R=415
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C
P=435, Q=825, R=415
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D
P=435, Q=825, R=415
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Solution

The correct option is A $$P=\frac { 4 }{ 35 } ,\ Q=-\frac { 8 }{ 25 } ,\ R=\frac { 4 }{ 15 } $$
Take $$1+x^{5/2}=t^2$$

$$\Rightarrow \dfrac{5}{2}x^{3/2}dx=2t \space dt$$

Rewrite the given integral as

$$\Rightarrow \int{x^5(1+x^{5/2})^{1/2}}x^{3/2}dx$$

Substitutng $$t$$ we get

$$\Rightarrow \dfrac{4}{5}\int(t^2-1)^2.t^2 dt$$

$$\Rightarrow \dfrac{4}{5}\int(t^4-2t^2+1).t^2 dt$$

$$\Rightarrow \dfrac{4}{5}\int(t^6-2t^4+t^2) dt$$

$$\Rightarrow \dfrac{4}{5} (\dfrac{t^7}{7}-\dfrac{2t^5}{5}+\dfrac{t^3}{3})+C$$

$$\Rightarrow \dfrac{4t^7}{35}-\dfrac{8t^5}{25}+\dfrac{4t^3}{15}+C$$

$$\Rightarrow \dfrac{4}{35}(1+x^{5/2)^{7/2}}-\dfrac{8}{25}(1+x^{5/2})^{5/2}+\dfrac{4}{15}(1+x^{5/2})^{3/2}+C$$

Comparing this with given expression we get

$$P=\dfrac{4}{35}, Q=\dfrac{-8}{25}, R=\dfrac{4}{15}$$

Mathematics

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