Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
बेरीज
Advertisements
उत्तर
`int_1^2 (x "d"x)/(x^2 + 1) = 1/2 int_1^2 (2xdx)/(x^2 + 1)`
= `1/2 int_1^2 ("d"(x^2 + 1))/(x^2 + 1)`
=`1/2 [log|x^2 + 1|]_1^2`
= `1/2 [log|2^2 + 1| - log|1^2+ 1|]`
= `1/2 [log 5 - log 2]`
= `1/2 log[5/2]` .......`{"Using" log "a" - log "b" = log ("a"/"b")}`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]
\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]
\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]
\[\int\limits_0^{15} \left[ x \right] dx .\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
`int x^3/(x + 1)` is equal to ______.
