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  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 68 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 68 maths

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Answer: \frac{2}{9}\left(1+x^{3}\right)^{\frac{3}{2}}-\frac{2}{3}\left(1+x^{3}\right)^{\frac{1}{2}}+c

Hint: Use substitution method to solve this integral

Given: \int \frac{x^{5}}{\sqrt{1+x^{3}}} d x

Solution:

        \begin{aligned} &\text { Let } I=\int \frac{x^{5}}{\sqrt{1+x^{3}}} d x \\ &\text { Put } 1+x^{3}=t^{2} \Rightarrow 3 x^{2} d x=2 t \; d t \\ &\Rightarrow d x=\frac{2 t}{3 x^{2}} d t \text { then } \end{aligned}

        I=\int \frac{x^{5}}{\sqrt{t^{2}}} \frac{2 t \; d t}{3 x^{2}}=\int \frac{2}{3} \frac{x^{3} \cdot t}{t} d t=\frac{2}{3} \int x^{3} d t

           =\frac{2}{3} \int\left(t^{2}-1\right) \mathrm{dt} \quad\left[\because 1+x^{3}=t^{2} \Rightarrow t^{2}-1=x^{3}\right]

           =\frac{2}{3} \int t^{2} d t-\frac{2}{3} \int 1 d t=\frac{2}{3} \int t^{2} d t-\frac{2}{3} \int t^{0} d t

           =\frac{2}{3} \frac{t^{2+1}}{2+1}-\frac{2}{3} \frac{t^{0+1}}{0+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

           =\frac{2}{3} \cdot \frac{t^{3}}{3}-\frac{2}{3} t+c

            =\frac{2}{9}\left(1+x^{3}\right)^{\frac{3}{2}}-\frac{2}{3}\left(1+x^{3}\right)^{\frac{1}{2}}+\mathrm{c} \quad\left[\because t^{2}=1+x^{3} \Rightarrow t=\sqrt{1+x^{3}}\right]

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