Advertisements
Advertisements
Question
\[\int\left( x^e + e^x + e^e \right) dx\]
Sum
Advertisements
Solution
\[\int\left( x^e + e^x + e^e \right)dx\]
\[ = \int x^e dx + \int e^x dx + e^e \int1dx\]
\[ = \frac{x^{e + 1}}{e + 1} + e^x + x \cdot e^e + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int\frac{e^\sqrt{x} \cos \left( e^\sqrt{x} \right)}{\sqrt{x}} dx\]
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{\cos x}{\sin^2 x + 4 \sin x + 5} dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]
\[\int x^3 \cos x^2 dx\]
\[\int\frac{x^2 \tan^{- 1} x}{1 + x^2} \text{ dx }\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x + 1}{x \left( 1 + x e^x \right)} dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int \tan^5 x\ dx\]
\[\int \cot^5 x\ dx\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{\cos x}{\frac{1}{4} - \cos^2 x} \text{ dx }\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
