Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Find the integral: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$
✓
Answer
Let, I = $∫\frac{(x+2)}{(2x^2+6x+5)}dx$ we express $x+2=\mathrm{A} \frac{d}{d x}\left(2 x^{2}+6 x+5\right)+\mathrm{B}$ = A(4x + 6) + B Equating the coefficients of x and the constant terms from both sides, we get 4A = 1 and 6A + B = 2 or $A=\frac{1}{4}$ and $B=\frac{1}{2}$ Therefore, $\int \frac{x+2}{2 x^{2}+6 x+5}=\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}$ = $\frac{1}{4} \mathrm{I}_{1}+\frac{1}{2} \mathrm{I}_{2}$ .....(i) In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt Therefore, $I_{1}=\int \frac{d t}{t}=\log |t|+C_{1}$ = log |2x2 + 6x + 5| + C .....(ii) and $I_{2}=\int \frac{d x}{2 x^{2}+6 x+5}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{5}{2}}$ = $\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}$ Put $x+\frac{3}{2}=t$, so that dx = dt, we get $I_{2}=\frac{1}{2} \int \frac{d t}{t^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2 \times \frac{1}{2}} \tan ^{-1} 2 t+C_{2}$ $\tan ^{-1} 2\left(x+\frac{3}{2}\right)+\mathrm{C}_{2}=\tan ^{-1}(2 x+3)+\mathrm{C}_{2}$ ......(iii) Using (ii) and (iii) in (i), we get $\int \frac{x+2}{2 x^{2}+6 x+5} d x=\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+\mathrm{C}$ where, $C=\frac{C_{1}}{4}+\frac{C_{2}}{2}$
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