Get Answers to all your Questions

header-bg qa

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 1 maths Textbook Solution.

Answers (1)

Answer:\sqrt{x^{2}+6 x+10}-3\left[\log \left|(x+3)+\sqrt{x^{2}+6 x+10}\right|\right]+c

Given: \int \frac{x}{\sqrt{x^{2}+6 x+10}} d x

Hint: Simplify it and integrate it with the help of formula.

Solution: Let I=\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x

                     =\int \frac{x+3-3}{\sqrt{x^{2}+6 x+10}} d x                                                    ......................Adding +3 and -3 in numerator

                  I=\frac{1}{2} \int \frac{2(x+3)}{\sqrt{x^{2}+6 x+10}} d x-3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}}

         I=I_{1}-I_{2}          ..................(1)


                           \begin{aligned} &I_{1}=\frac{1}{2} \int \frac{2(x+3)}{\sqrt{x^{2}+6 x+10}} d x \\ &I_{2}=3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}} \end{aligned}

                           First solve I_{1}


                          \begin{aligned} &x^{2}+6 x+10=y \\ &(2 x+6) d x=d y \end{aligned}                            .........(2)

         I_{1}=\frac{1}{2} \int \frac{d y}{\sqrt{y}}=\frac{1}{2} \int(y)^{-\frac{1}{2}} d y


     I_{1}=\sqrt{y+c}                                From equation (2)

    I_{1}=\sqrt{x^{2}+6 x+10}+c

  Now I_{2}=3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}}

  x^{2}+6 x+10=x^{2}+6 x+9-9+10

=\left ( x+3 \right )^{2}+1

I_{2}=3 \int \frac{d x}{\sqrt{(x+3)^{2}+1}}

I_{2}=3 \log \left|x+3+\sqrt{x^{2}+6 x+10}\right|+c

\left[\int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left|x+\sqrt{x^{2}+a^{2}}\right|\right]

Now, putting value of I_{1} & I_{2} in equation 1

              I=\sqrt{x^{2}+6 x+10}-3\left[\log \left|(x+3)+\sqrt{x^{2}+6 x+10}\right|\right]+c


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support