Ncert Solutions Maths class 12th

$\frac{2x}{x^2+3x+2}=\frac{2x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

2x = A(x + 2) + B(x + 1).

= (A + B) x + (2A + B).

Comparing the coefficients we get,

A + B = 2 ---(1)

2A + B = 0 ---- (2).

So, Eqⁿ (2) - (1),

2A + B - (A + B) = 0 - 2

Þ A = - 2

And B = 2 - A = 2 - (-2) = 2 + 2 = 4

B = 4

$\begin{array}{l}So,\frac{2x}{(x+1)(x+2)}=\frac{-2}{x+1}+\frac{4}{x+2}.\\ \therefore {\displaystyle \int \frac{2xdx}{(x+1)(x+2)}}=-2{\displaystyle \int \frac{dx}{x+1}}+4{\displaystyle \int \frac{dx}{x+2}}\\ =-2log|x+1|+4log|x+2|+c\\ =4log|x+2|-2log|x+1|+c\end{array}$

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$\begin{array}{l}So,\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}+\frac{(-5)}{(x-2)}+\frac{4}{(x-3)}\\ Then,{\displaystyle \int \frac{3x-1dx}{(x-1)(x-2)(x-3)}}={\displaystyle \int \frac{dx}{x-1}}-5{\displaystyle \int \frac{dx}{x-2}}+4{\displaystyle \int \frac{dx}{x-3}}.\\ =log|x-1|-5log|x-2|+4log|x-3|+c\end{array}$

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$\begin{array}{l}I=\int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(x^2+2x+1)+1}\\ =\int \frac{dx}{{(x+1)}^2+1}dx\\ \begin{array}{l}= \left[ta{n}^{–1}\left(x+1\right)\right] +C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Hence, the correct Answer is \left(B\right).\hfill \end{array}\end{array}$

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$\begin{array}{l}Let\left(x+3\right)=A\frac{d}{dx}\left(x^2–2x–5\right)+B\\ \Rightarrow \left(x+3\right)=A\left(2x–2\right)+B \\ Equatingthecoefficientofxandconstanttermonbothside,weget\\ Weget 2A=1 and –2A+B=3 \\ A= \frac{1}{2} B=4 \\ Therefore,\left(x+3\right)=\frac{1}{2}(2x-2)+4-----(1)\\ I=\int \frac{x+3}{x^2-2x-5}dx=\int \frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5}dx\\ =\frac{1}{2}\int \frac{2x-2}{x^2-2x-5}dx+4\int \frac{1}{x^2-2x+5}dx\\ I=\int \frac{x+3}{x^2-2x-5}dx=\frac{1}{2}{I}_1+4I_2\\ I_1=\int \frac{2x-2}{x^2-2x-5}dx\\ \begin{array}{ll}Let{x}^2– 2x– 5 =t\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow \left(2x–2\right)dx=dt\hfill \end{array}\\ I_1=\int \frac{dt}{t}=log\left|t\right|=log\left|x^2-2x-5\right|-----(2)\\ I_2=\int \frac{1}{x^2-2x-5}dx=\int \frac{1}{\left(x^2-2x+1\right)-6}dx=\int \frac{1}{{(x-1)}^2-(\surd 6){}^2}dx\\ =\frac{1}{\mathrm{2\surd 6}}log\left(\frac{x-1-\surd 6}{x-1+\surd 6}\right)-----(3)\\ Substituting\left(2\right)and\left(3\right)in\left(1\right),weget\\ I=\frac{1}{2}log\left|x^2-2x-5\right|+\frac{4}{2}log\left|\frac{x-1-\surd 6}{x-1+\surd 6}\right|+C\end{array}$$\begin{array}{l}\end{array}$

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