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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$
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Answer

Let $\text{I}=\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$ Rewriting the numerator we have, $5\text{x}-2=\text{A}\frac{\text{d}}{\text{dx}}\big(1+2\text{x}+3\text{x}^2\big)+\text{B}$ $\Rightarrow5\text{x}-2=\text{A}(2+6\text{x})+\text{B}$ $\Rightarrow5\text{x}-2=6\text{xa}+2\text{A}+\text{B}$ Comparing the coefficient, we have, $6\text{A}=5$ and $2\text{A}+\text{B}=-2$ $\Rightarrow\text{A}=\frac{5}{6}$ Substituting the value of A in 2A + B = -2, we have, $2\times\frac{5}{6}+\text{B}=-2$ $\Rightarrow\frac{10}{6}+\text{B}=-2$ $\Rightarrow\text{B}=-2-\frac{10}{6}$ $\Rightarrow\text{B}=\frac{-12-10}{6}$ $\Rightarrow\text{B}=\frac{-22}{6}$ $\Rightarrow\text{B}=\frac{-11}{3}$ $5\text{x}-2=\frac{5}{6}(2+6\text{x})-\frac{11}{3}$Thus, $\text{I}=\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$ becomes,
$\text{I}=\int\frac{\big[\frac{5}{6}(2+6\text{x})-\frac{11}{3}\big]}{3\text{x}^2+2\text{x}+1}\text{ dx}$ $=\frac{5}{6}\int\frac{(2+6\text{x})}{3\text{x}^2+2\text{x}+1}\text{ dx}-\frac{11}{3}\int\frac{\text{dx}}{3\text{x}^2+2\text{x}+1}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{3\times3}\int\frac{\text{dx}}{\text{x}^2+\frac{2}{3}\text{x}+\frac{1}{3}}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\int\frac{\text{dx}}{\text{x}^2+\frac{2}{3}\text{x}+\big(\frac{4}{3}\big)^2+\frac{1}{3}-\big(\frac{4}{3}\big)^2}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\int\frac{\text{dx}}{\big(\text{x}+\frac{1}{3}\big)^2+\Big(\frac{\sqrt2}{3}\Big)^2}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\times\frac{1}{\frac{\sqrt2}{3}}\tan^{-1}\Bigg[\frac{\big(\text{x}+\frac{1}{3}\big)}{\frac{\sqrt2}{3}}\Bigg]+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\times\frac{3}{\sqrt2}\tan^{-1}\Bigg[\frac{\big(\frac{3\text{x}+1}{3}\big)}{\frac{\sqrt2}{3}}\Bigg]+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{3\sqrt2}\tan^{-1}\Big[\frac{3\text{x}+1}{\sqrt2}\Big]+\text{C}$

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