Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[I = \int\frac{x + 5}{3 x^2 + 13x - 10}dx\]
\[ = \int\frac{x + 5}{3 x^2 + 15x - 2x - 10}dx\]
\[ = \int\frac{x + 5}{3x\left( x + 5 \right) - 2\left( x + 5 \right)}dx\]
\[ = \int\frac{x + 5}{\left( 3x - 2 \right)\left( x + 5 \right)}dx\]
\[= \int\frac{x + 5}{(3x - 2)(x + 5)}dx\]
\[ = \int\frac{1}{3x - 2}dx\]
\[ \therefore I = \frac{1}{3}\text{ ln }\left| 3x - 2 \right| + c\]
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^3dx/(9+x^2)`
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
Evaluate the following integrals:
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
