Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
योग
Advertisements
उत्तर
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x = int_(-1)^1 ("d"(x^2 + 3x + 7))/(x^2 + 3x + 7)`
= `[log|x^2 + 3x + 7|]_(-1)^1`
= `log|1 + 3 + 7| - log|1 - 3 + 7|`
= `log 11 - log 5`
= `log [11/5]`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx\]
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\]
\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]
\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]
\[\int_0^\pi \cos x\left| \cos x \right|dx\]
\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]
\[\int\limits_0^2 e^x dx\]
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
