Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\frac{e^x dx}{e^{2x} + 5 e^x + 6}\]
\[\text{ let } e^x = t\]
\[ \Rightarrow e^x \text{ dx }= dt\]
\[Now, \int\frac{e^x dx}{e^{2x} + 5 e^x + 6}\]
\[ = \int\frac{dt}{t^2 + 5t + 6}\]
\[ = \int\frac{dt}{t^2 + 5t + \left( \frac{5}{2} \right)^2 - \left( \frac{5}{2} \right)^2 + 6}\]
\[ = \int\frac{dt}{\left( t + \frac{5}{2} \right)^2 - \frac{25}{4} + 6}\]
\[ = \int\frac{dt}{\left( t + \frac{5}{2} \right)^2 - \frac{25 + 24}{4}}\]
\[ = \int\frac{dt}{\left( t + \frac{5}{2} \right)^2 - \left( \frac{1}{2} \right)^2}\]
\[ = \frac{1}{2 \times \frac{1}{2}} \log \left| \frac{t + \frac{5}{2} - \frac{1}{2}}{t + \frac{5}{2} + \frac{1}{2}} \right| + C\]
\[ = \text{ log }\left| \frac{t + 2}{t + 3} \right| + C\]
\[ = \text{ log} \left| \frac{e^x + 2}{e^x + 3} \right| + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
` = ∫ root (3){ cos^2 x} sin x dx `
If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
