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Find \(\int \frac{(x+3)}{2x^2+6x+7} dx\).(a) \(\frac{1}{4} log(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}\frac{2x+3}{2\sqrt{2}}\right )+C\)(b) \(\frac{1}{4} log(2x^2+6x+7) – \frac{3}{4} (\frac{1}{\sqrt{2}} tan^{-1}\frac{2x+3}{2\sqrt{2}} )+C\)(c) \(log(2x^2+6x+7) + \left (tan^{-1}\frac{2x+3}{2√2}\right )+C\)(d) –\(log(2x^2+6x+7) – \frac{3}{4} \left (\frac{1}{√2} tan^{-1}\frac{2x+3}{2√2}\right )+C\)The question was asked in an online quiz.The question is from Integrals of Some Particular Functions in portion Integrals of Mathematics – Class 12 |
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Answer» RIGHT answer is (a) \(\frac{1}{4} log(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}\frac{2x+3}{2\sqrt{2}}\right )+C\) EXPLANATION: We can express x+3=A \(\frac{d}{DX}\) (2x^2+6x+7)+B x+3=A(4x+6)+B x+3=4Ax+(6A+B) COMPARING the coefficients, we get 4A=1 ⇒A=1/4 6A+B=3 ⇒B=3/2 \(\int \frac{x+3}{2x^2+6x+7} dx=\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx+\frac{3}{2} \int \frac{1}{2x^2+6x+7} dx\) Let 2x^2+6x+7=t (4x+6)dx=dt \(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} \int \frac{dt}{t}=\frac{1}{4} logt\) Replacing t with (2x^2+6x+7) \(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} log(2x^2+6x+7)\) \(\frac{3}{2} \int \frac{1}{2x^2+6x+7} dx=\frac{3}{2} \int \frac{1}{2(x^2+3x+\frac{7}{2})} dx=\frac{3}{4} \int \frac{1}{(x+\frac{3}{2})^2+2} dx\) =\(\frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}\frac{2x+3}{2\sqrt{2}} \right )\) ∴\(\int \frac{x+3}{2x^2+6x+7} dx=\frac{1}{4} log(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}\frac{2x+3}{2√2} \right )+C\) |
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