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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.10 Question 10

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Answer:  3 x^{\frac{1}{3}}+3 \log \left|x^{\frac{1}{3}}-1\right|+c

Hint: Use substitution method to solve this type of integral

Given:  \int \frac{1}{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)} d x

Solution: let  I=\int \frac{1}{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)} d x

Substitute  x=t^{3} \Rightarrow d x=3 t^{2} d t   then

\begin{aligned} I &=I=\int \frac{1}{t(t-1)} 3 t^{2} d t \quad\left(\because x^{\frac{1}{3}}=t\right) \\ & \end{aligned}

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

=3 \int \frac{t+1-1}{t-1} d t=3 \int\left(\frac{t-1}{t-1}+\frac{1}{t-1}\right)dt

\begin{aligned} &=3 \frac{t^{0+1}}{0+1}+3 \log |t-1|+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}

=3 t+3 \log |t-1|+c \\

\therefore I=3 x^{\frac{1}{3}}+3 \log \left|x^{\frac{1}{3}}-1\right|+c

 

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