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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.10 Question 8

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Answer:  \frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c

Hint: Use substitution method to solve this type of integral

Given:  \int x(1-x)^{23} d x

Solution: let  I=\int x(1-x)^{23} d x

Substitute 1-x=t \Rightarrow-d x=d t  then

\begin{aligned} &I=\int(1-t) \cdot t^{23}(-d t) \qquad(\because x=1-t) \\ & \end{aligned}

\Rightarrow I=-\int(1-t) \cdot t^{23} d t=-\int t^{23}-t . t^{23} d t \\

\Rightarrow I=-\int\left(t^{23}-t^{24}\right) d t=-\int t^{23} d t+\int t^{24} d t

\begin{aligned} &\Rightarrow I=-\frac{t^{23+1}}{23+1}+\frac{t^{24+1}}{24+1}+c \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}

\Rightarrow I=\frac{t^{25}}{25}-\frac{t^{24}}{24}+c \\

\therefore I=\frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c

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