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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 12 Maths Textbook Solution.

Answers (1)

Answer: \frac{5}{6} \ln \left|3 x^{2}+2 x+1\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1} \frac{3 x+1}{\sqrt{2}}+c

Hint: Find value of M and N

Given: \int \frac{5 x-2}{1+2 x+3 x^{2}} d x

Solution: I=\int \frac{5 x-2}{1+2 x+3 x^{2}} d x

                \begin{aligned} &5 x-2=M \frac{d}{d x}\left(1+2 x+3 x^{2}\right)+N \\ &5 x-2=M(2+6 x)+N \\ &5 x-2=2 M+6 M x+N \end{aligned}

On comparing,

                \begin{aligned} &6 M=5 \Rightarrow M=\frac{5}{6} \\ &2 M+N=-2 \Rightarrow 2\left(\frac{5}{6}\right)+N=-2 \\ &\Rightarrow \frac{5}{3}+N=-2 \Rightarrow N=-2-\frac{5}{3}=-\frac{11}{3} \end{aligned}

I=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^{2}} d x-\frac{11}{3} \int \frac{d x}{1+2 x+3 x^{2}}                ................................(1)

\begin{aligned} &I_{1}=\int \frac{2+6 x}{1+2 x+3 x^{2}} d x \\ &\text { let, } \quad t=1+2 x+3 x^{2} \\ &\Rightarrow 2+6 x=\frac{d t}{d x} \\ &\Rightarrow(2+6 x) d x=d t \end{aligned}

                                                                                                        d x=\frac{d t}{2+6 x}

\begin{aligned} &I_{1}=\int \frac{d t}{t}=\log |t|+c \\ &I_{2}=\int \frac{d x}{1+2 x+3 x^{2}} \end{aligned}

\begin{aligned} &=3\left(x^{2}+2 \cdot x \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2}+\frac{1}{3}\right) \\ &=3\left(\left(x+\frac{1}{3}\right)^{2}+\frac{1}{3}-\frac{1}{9}\right) \\ &=3\left(\left(x+\frac{1}{3}\right)^{2}+\frac{3-1}{9}\right) \\ &=3\left(\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right) \end{aligned}

From eq 1
I=\frac{5}{6} \int \frac{d t}{t}-\frac{11}{3} \int \frac{d x}{3\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}}

                   =\frac{5}{6} \log |t|-\frac{11}{9} \int \frac{d x}{\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}}                                                                \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c

                  =\frac{5}{6} \log \left|3 x^{2}+2 x+1\right|-\frac{11}{9} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1} \frac{\left(x+\frac{1}{3}\right)}{\frac{\sqrt{2}}{3}}

                I=\frac{5}{6} \log \left|3 x^{2}+2 x+1\right|-\frac{11}{2 \sqrt{3}} \tan ^{-1} \frac{(3 x+1)}{\sqrt{2}}+c

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