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  • need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 23

need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 23

Answers (1)

best_answer

Answer: 2 \sqrt{x}-2 \log |\sqrt{x}+1|+c

Hint:Use substitution method to solve this integral.

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

Solution:

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

        \begin{aligned} I &=\int \frac{1}{1+\sqrt{t^{2}}} 2 t d t=\int \frac{2 t}{1+t} d t \\ &=2 \int \frac{t}{1+t} d t=2 \int \frac{1+t-1}{1+t} d t \end{aligned}

            \begin{aligned} &=2 \int \frac{(1+t)-1}{1+t} d t=2 \int\left\{\frac{1+t}{1+t}-\frac{1}{1+t}\right\} d t \\ &=2 \int\left\{1-\frac{1}{1+t}\right\} d t=2 \int 1 . d t-2 \int \frac{1}{1+t} d t \end{aligned}

            =2 \int 1 . d t-2 \int \frac{1}{1+t} d t                            ......(i)                

        \text { Now } 2 \int 1 . d t=2 \frac{t^{0+1}}{0+1}+c_{1} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

            =2 t+c_{1}                                                          ........(ii)

        \begin{aligned} &\text { and } 2 \int \frac{1}{1+t} d t \\ &\text { Put } 1+t=p \Rightarrow d t=d p \text { then } \end{aligned}

        2 \int \frac{1}{1+t} d t=2 \int \frac{1}{p} d p=2 \log |p|+c_{2}

            =2 \log |t+1|+c_{2}                                        .........(iii)

        Putting the values of equation (ii) and (iii) in (i) then

        \begin{aligned} &I=2 t+c_{1}-\left(2 \log |t+1|+c_{2}\right) \\ &\Rightarrow I=2 t+c_{1}-2 \log |t+1|-c_{2} \\ &\therefore I=2 \sqrt{x}-2 \log |\sqrt{x}+1|+c_{1}-c_{2} \end{aligned}

        \therefore I=2 \sqrt{x}-2 \log |\sqrt{x}+1|+c \quad\left[\begin{array}{c} \because t^{2}=x \Rightarrow t=\sqrt{x} \\ \text { and } c=c_{1}-c_{2} \end{array}\right]

                                                                                      

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