Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
बेरीज
Advertisements
उत्तर
= `int_1^"e" (1 + logx)^-3/x "d"x`
= `[("f"(x)^(-3 + 1))/(-3 + 1)]_1^"e"`
= `[(1 + log x)^-2/-2]_1^"e"`
= `- 1/2 [[1 + log x]^-2]_1^"e"`
= `- 1/2 [(1 + log "e")^-2 (1 + log 1)^-2]`
= - 1/2 [(1 + 1)^-2 - (1)^-2]`
= `- 1/2 [1/(2)^2 - 1/(1)^2]`
= `- 1/2[1/4 - 1]`
= `-1/2[(1 - 4)/4]`
= `- 1/2[(-3)/4]`
= `3/8`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]
\[\int\limits_0^2 \left( x + 3 \right) dx\]
\[\int\limits_2^3 x^2 dx\]
\[\int\limits_1^e \log x\ dx =\]
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
