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The value of \int \frac{d x}{(1+x)^{1 / 2}-(1+x)^{1 / 3}} is 

Option: 1

2 \lambda^{1 / 2}+3 \lambda^{1 / 3}+6 \lambda^{1 / 6}+6 \ln \left|\lambda^{1 / 6}-1\right|+c


Option: 2

2 \lambda^{1 / 2}-3 \lambda^{1 / 3}+6 \lambda^{1 / 6}+6 \ln \left|\lambda^{1 / 6}-1\right|+c


Option: 3

2 \lambda^{1 / 2}+3 \lambda^{1 / 3}-6 \lambda^{1 / 6}+6 \ln \left|\lambda^{1 / 6}-1\right|+c


Option: 4

2 \lambda^{1 / 2}+3 \lambda^{1 / 3}+6 \lambda^{1 / 6}-6 \ln \left|\lambda^{1 / 6}-1\right|+c


Answers (1)

best_answer

Where \lambda=(1+x)

Let                 I=\int \frac{d x}{(1+x)^{1 / 2}-(1+x)^{1 / 3}}

Put                1+x=t^6

\Rightarrow \: \: \: \: \quad d x=6 t^5 d t

Then,

       \begin{aligned} & I=\int \frac{6 t^5 d t}{\left(t^3-t^2\right)}=6 \int \frac{t^3}{(t-1)} d t \\ \\& \: \: =6 \int \frac{\left(t^3-1\right)+1}{(t-1)} d t \\ \\& \: \: \: =6 \int\left(t^2+t+1+\frac{1}{t-1}\right) d t \end{aligned}

            \begin{aligned} & =6\left\{\frac{t^3}{3}+\frac{t^2}{2}+t+\ln |t-1|\right\}+c \\ \\& =2(1+x)^{1 / 2}+3(1+x)^{1 / 3}+6(1+x)^{1 / 6} \\ \\&\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad+6 \ln \left|(1+x)^{1 / 6}-1\right|+c \end{aligned}

Posted by

Divya Prakash Singh

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