Question

Evaluate the integration of 1/(2x^2+3x+1) dx​​

Answer 1

Step-by-step explanation:

$\implies \int \frac{1 }{2x {}^{2} + 3x + 1 } \small{ \mathsf{dx}} \\ \\ \implies\int \frac{ \frac{1}{2} }{x {}^{2} + \frac{3}{2}x + \frac{ 1}{2} } \small{ \mathsf{dx}} \\ \\ \implies \small{\frac{1 }{2}}\int \frac{ 1}{x {}^{2} + \small{2} ( \frac{3}{4})(x) + (\frac{3}{4} ) {}^{2} - (\frac{3}{4}) {}^{2} + \frac{ 1}{2} } \small{ \mathsf{dx}} \\ \\ \implies \small{ \frac{1 }{2}}\int \frac{1}{(x + \frac{3}{4} ) {}^{2} - (\frac{3}{4}) {}^{2} + \frac{ 1}{2} } \small{ \mathsf{dx}} \\ \\ \implies \small{\frac{1 }{2}}\int \frac{1}{(x + \frac{3}{4} ) {}^{2} - \frac{ 1}{16} } \small{ \mathsf{dx}}$

$\sf{ \bold{\small{let \: x + \frac{3}{4} = t\: \: \: \: \rightarrow dx = dt}}}$

$\implies \small{\frac{1 }{2}}\int \frac{1}{t {}^{2} -( \frac{ 1}{4}) {}^{2} } \small{ \mathsf{dt}} \\ \\\implies \small{\frac{1 }{2}} \times \frac{1}{2( \frac{1}{4}) } \mathsf{ \times ln \large{|} \small{\frac{t - \frac{1}{4} }{t + \frac{1}{4} } }} \large{|} \small{ \sf{ \: + C} } \\ \\\implies \mathsf{ \frac{1}{2} \times 2 \times \: ln \large{|} \small{\frac{t - \frac{1}{4} }{t + \frac{1}{4} } }} \large{|} \: \:, \small{ \sf{ \: + C} }\\ \\ \implies\mathsf{ ln \large{|} \small{\frac{x + \frac{3}{4} - \frac{1}{4} }{x + \frac{3}{4} + \frac{1}{4} } }} \large{|} \small{ \sf{ \: + C} }\\ \\ \implies\mathsf{ ln \large{|} \small{\frac{2x + 1}{2x + 2} }} \large{|} \small{ \sf{ \: + C} } \\ \\ \implies \sf{ln |2x + 1|\: - ln|2x + 2 |} + C$

$\sf{Notice \: that\: ln|2x + 2| + C = ln|x + 1| + ln2 + C} \\ \sf{= ln|x + 1| + C_1 , \: \:as\: ln2\: is \:constant.}$

$\implies\bold{\sf{ln |2x + 1|\: - ln|x + 1|} + C}$