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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18 .19 Question 11 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18 .19 Question 11 Maths Textbook Solution.

Answers (1)

Answer:  \frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+\frac{1}{2} \tan ^{-1}(2 x+3)+c

Hint: Multiply and Divide by 4

Given: \int \frac{x+2}{2 x^{2}+6 x+5} d x

Solution:Multiply and divide by 4

              \frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{2}{4} \int \frac{d x}{2 x^{2}+6 x+5}  .........(1)

              \begin{aligned} &I_{1}=\int \frac{4 x+6}{2 x^{2}+6 x+5} d x \\ &2 x^{2}+6 x+5=t \\ &(4 x+6) d x=d t \\ &I_{1}=\int \frac{d t}{t}=\ln t+c=\ln \left(2 x^{2}+6 x+5\right)+c_{1} \end{aligned}

              \begin{aligned} &I_{2}=\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5} \\ &I_{2}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{9}{4}+\frac{5}{2}-\frac{9}{4}} \end{aligned}

               \frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\frac{1}{4}}

        \begin{aligned} &\left(x+\frac{3}{2}\right)=u \Rightarrow 1 d x=d u \\ &I_{2}=\frac{1}{2} \int \frac{d u}{u^{2}+\left(\frac{1}{2}\right)^{2}} \\ &\frac{1}{2}\left[\frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{u}{\frac{1}{2}}\right)\right]+c_{2}\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}\right. \\ &\begin{array}{l} \tan ^{-1}\left[2 \tan ^{-1}2\left(x+\frac{3}{2}\right)\right]+c_{2} \\ I_{2}=\tan ^{-1}(2 x+3)+c_{2} \\ I=\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+c_{1}+\frac{1}{2} \tan ^{-1}(2 x+3)+c_{2} \\ I=\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+\frac{1}{2} \tan ^{-1}(2 x+3)+c \end{array} \end{aligned}

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