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  • Please Solve R D Sharma class 12 Chapter 19 Definite Integrals Exercise 19 Point 1 Question 65 Maths textbook Solution

Please Solve R.D. Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 65 Maths textbook Solution.

Answers (2)

Answer: \frac{3}{8}log 3

Hint: Use indefinite formula then put the limit to solve this integral

Given: \int_{0}^{1}xlog\left ( 1+2x \right )dx

Solution: \int_{0}^{1}xlog\left ( 1+2x \right )dx

Apply integration by parts, we get

=\left[\log (1+2 x) \int x d x\right]_{0}^{1}-\int_{0}^{1}\left\{\frac{d}{d x} \log (1+2 x) \int x d x\right\} d x                                                \left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \frac{d}{d x}(\log (a x+b))=\frac{1}{a x+b} a \end{array}\right]

=\left[\log (1+2 x) \frac{x^{1+1}}{1+1}\right]_{0}^{1}-\int_{0}^{1}\left(\frac{1}{1+2 x} 2 \frac{x^{1+1}}{1+1}\right) d x                          

\begin{aligned} &=\left[\log (1+2 x) \frac{x^{2}}{2}\right]_{0}^{1}-2 \int_{0}^{1} \frac{1}{2 x+1} \frac{x^{2}}{2} d x \\ &=\left[\log (1+2(1)) \frac{1^{2}}{2}-\log (1+2(0)) \frac{0^{2}}{2}\right]-\frac{1}{2} \int_{0}^{1} \frac{2 x^{2}}{2 x+1} d x \\ &=\left[\frac{\log 3}{2}-\log 1 \times 0\right]-\frac{1}{2} \int_{0}^{1} \frac{2 x^{2}+x-x}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} \frac{x(2 x+1)-x}{2 x+1} d x \end{aligned}

\begin{aligned} &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1}\left(\frac{x(2 x+1)}{2 x+1}-\frac{x}{2 x+1}\right) d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} x d x+\frac{1}{2} \int_{0}^{1} \frac{2 x+1-1}{2(2 x+1)} d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} x d x+\frac{1}{2} \int_{0}^{1}\left(\frac{2 x+1}{(2 x+1)}-\frac{1}{2 x+1}\right) d x \\ &=\frac{\log 3}{2}-\frac{1}{2}\left[\frac{x^{1+1}}{1+1}\right]_{0}^{1}+\frac{1}{4} \int_{0}^{1} 1 d x-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \end{aligned}                                                        \left [ x^{n}dx=\frac{x^{n+1}}{n+1} \right ]

\begin{aligned} &=\frac{\log 3}{2}-\frac{1}{2}\left[\frac{x^{2}}{2}\right]_{0}^{1}+\frac{1}{4}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{1}-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{4}\left[1^{2}-0^{2}\right]+\frac{1}{4}[1-0]-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \end{aligned}

Put 2x+1=t\Rightarrow 2dx=dt\Rightarrow dx=\frac{dt}{2}

When x=0 then t=2 and

When x=1 then t=3

                                                  

\begin{aligned} &=\frac{\log 3}{2}-\frac{1}{4} \int_{1}^{3} \frac{1}{t} \frac{d t}{2} \\ &=\frac{\log 3}{2}-\frac{1}{8}[\log |t|]_{1}^{3} \\ &=\frac{\log 3}{2}-\frac{1}{8}[\log 3-\log 1] \\ &=\frac{\log 3}{2}-\frac{1}{8}[\log 3]+0 \\ &=\log 3\left(\frac{1}{2}-\frac{1}{8}\right) \\ &=\log 3\left(\frac{4-1}{8}\right) \\ &=\frac{3}{8} \log 3 \end{aligned}

 

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Answer:\frac{3}{8}log\: 3

Hint: Use indefinite formula then put the limit to solve this integral

Given:\int_{0}^{1}xlog\left ( 1+2x \right )dx

Solution:\int_{0}^{1}xlog\left ( 1+2x \right )dx

Apply integration by parts, we get

=\left[\log (1+2 x) \int x d x\right]_{0}^{1}-\int_{0}^{1}\left\{\frac{d}{d x} \log (1+2 x) \int x d x\right\} d x                            \left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \frac{d}{d x}(\log (a x+b))=\frac{1}{a x+b} a \end{array}\right]

=\left[\log (1+2 x) \frac{x^{1+1}}{1+1}\right]_{0}^{1}-\int_{0}^{1}\left(\frac{1}{1+2 x} 2 \frac{x^{1+1}}{1+1}\right) d x

\begin{aligned} &=\left[\log (1+2 x) \frac{x^{2}}{2}\right]_{0}^{1}-2 \int_{0}^{1} \frac{1}{2 x+1} \frac{x^{2}}{2} d x \\ &=\left[\log (1+2(1)) \frac{1^{2}}{2}-\log (1+2(0)) \frac{0^{2}}{2}\right]-\frac{1}{2} \int_{0}^{1} \frac{2 x^{2}}{2 x+1} d x \\ &=\left[\frac{\log 3}{2}-\log 1 \times 0\right]-\frac{1}{2} \int_{0}^{1} \frac{2 x^{2}+x-x}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} \frac{x(2 x+1)-x}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1}\left(\frac{x(2 x+1)}{2 x+1}-\frac{x}{2 x+1}\right) d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} x d x+\frac{1}{2} \int_{0}^{1} \frac{2 x+1-1}{2(2 x+1)} d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} x d x+\frac{1}{2} \int_{0}^{1}\left(\frac{2 x+1}{(2 x+1)}-\frac{1}{2 x+1}\right) d x \\ &=\frac{\log 3}{2}-\frac{1}{2}\left[\frac{x^{1+1}}{1+1}\right]_{0}^{1}+\frac{1}{4} \int_{0}^{1} 1 d x-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \end{aligned}

                                                                                                                                            

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