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# If $\int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=P{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{7}{2}}}+Q{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{7}{2}}}+R{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{7}{2}}}+C$ then P, Q and R are \begin{align} & a)P=\dfrac{4}{35},Q=-\dfrac{8}{25},R=\dfrac{4}{15} \\ & b)P=\dfrac{4}{35},Q=\dfrac{8}{25},R=\dfrac{4}{15} \\ & c)P=-\dfrac{4}{35},Q=-\dfrac{8}{25},R=\dfrac{4}{15} \\ & d)P=\dfrac{4}{35},Q=\dfrac{8}{25},R=-\dfrac{4}{15} \\ \end{align}

Last updated date: 24th Jul 2024
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Hint: Now consider the given integral .To solve the integral we will use a method of substitution. Hence we will first substitute $1+{{x}^{\dfrac{5}{2}}}={{t}^{2}}$ now differentiate the equation to find relation between dt and dx. Now we will write the whole integral with respect to t. Now we know that $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C$ Hence we will use this formula to expand the obtained integral. Now we will simplify the equation and compare the equation with the given equation. Hence we get the required values of P, Q and R.

Now consider the given integral $\int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}$ . We know that ${{a}^{m+n}}={{a}^{n}}{{a}^{m}}$ using this we get,
$\int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=\int{{{x}^{\dfrac{5}{2}}}{{x}^{\dfrac{3}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}....................\left( 1 \right)$
To solve this integral we will use a method of substitution.
Now let $1+{{x}^{\dfrac{5}{2}}}={{t}^{2}}$
Differentiating the above equation we get,
\begin{align} & \dfrac{5}{2}{{x}^{\dfrac{3}{2}}}dx=2tdt \\ & \Rightarrow {{x}^{\dfrac{3}{2}}}dx=\dfrac{4}{5}tdt \\ \end{align}
Now since $1+{{x}^{\dfrac{5}{2}}}={{t}^{2}}$ we have,
$\Rightarrow {{x}^{\dfrac{5}{2}}}={{t}^{2}}-1$
Squaring the above equation on both sides we get,
$\Rightarrow {{x}^{5}}={{\left( {{t}^{2}}-1 \right)}^{2}}$
Now substituting these values in equation (1) we will write the integral in terms of t,
$\Rightarrow \int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=\int{\dfrac{4}{5}{{\left( {{t}^{2}}-1 \right)}^{2}}{{t}^{2}}dt}$
Now simplifying the equation using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ we get,
\begin{align} & \Rightarrow \int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=\dfrac{4}{5}\int{\left( {{t}^{4}}-2{{t}^{2}}+1 \right){{t}^{2}}dt} \\ & \Rightarrow \int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=\dfrac{4}{5}\int{\left( {{t}^{6}}-2{{t}^{4}}+{{t}^{2}} \right)dt} \\ \end{align}
Now we know that $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C$ hence using this we get,
\begin{align} & \Rightarrow \dfrac{4}{5}\int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=\dfrac{4}{5}\left( \dfrac{{{t}^{7}}}{7}-2\dfrac{{{t}^{5}}}{5}+\dfrac{{{t}^{3}}}{3} \right)+C \\ & \Rightarrow \int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=\left( \dfrac{4{{t}^{7}}}{35}-\dfrac{8{{t}^{5}}}{25}+\dfrac{4{{t}^{3}}}{15} \right)+C \\ \end{align}
Now substituting the value of t we get,
Now let us find P, Q and R by comparing the above equation with the equation$\int{{{x}^{\dfrac{13}{2}}}{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{1}{2}}}dx}=P{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{7}{2}}}+Q{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{7}{2}}}+R{{\left( 1+{{x}^{\dfrac{5}{2}}} \right)}^{\dfrac{7}{2}}}+C$
Hence we get, $P=\dfrac{4}{35},Q=\dfrac{-8}{25}R=\dfrac{4}{15}$
Hence option a is the correct option.

Note: Now we always have to be careful when using substitution to solve integration. Also while replacing the variable also remember to replace the differential dx by differentiating the equation and find its relation to the variable dt. Note that whenever we have such indices in multiplication we can use the method of substitution in the above manner. Also in case of definite integral remember to change the limits as well.