Advertisements
Advertisements
Question
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
Sum
Advertisements
Solution
\[\text{Let I} = \int\frac{e^{3x}}{e^{3x} + 1}dx\]
\[\text{Putting }e^{3x} + 1 = t \]
\[ \Rightarrow 3 e^{3x} = \frac{dt}{dx}\]
\[ \Rightarrow dx = \frac{dt}{3 e^{3x}}\]
\[ \therefore I = \int\frac{e^{3x}}{3t\left( e^{3x} \right)}dt\]
\[ = \frac{1}{3}\int\frac{1}{t}dt\]
\[ = \frac{\text{ln }\left| t \right|}{3} + C\]
\[ = \frac{\text{ln} \left| e^{3x} + 1 \right|}{3} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
`int{sqrtx(ax^2+bx+c)}dx`
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]
` ∫ tan 2x tan 3x tan 5x dx `
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
\[\int \sin^4 x \cos^3 x \text{ dx }\]
\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text
{dx\]
\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int x e^{2x} \text{ dx }\]
\[\int\frac{\log \left( \log x \right)}{x} dx\]
\[\int x \cos^2 x\ dx\]
\[\int {cosec}^3 x\ dx\]
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int x \sin x \cos 2x\ dx\]
\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]
\[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 3 \right)} dx\]
\[\int\frac{1}{x^4 - 1} dx\]
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\frac{1}{e^x + 1} \text{ dx }\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
