Advertisements
Advertisements
Question
Evaluate:
Advertisements
Solution
\[\text{ Let I }= \int\left( \frac{x^2 + 4x}{x^3 + 6 x^2 + 5} \right) dx\]
\[\text{ Let x}^3 + 6 x^2 + 5 = t\]
\[ \Rightarrow \left( 3 x^2 + 12x \right) dx = dt\]
\[ \Rightarrow \left( x^2 + 4x \right) dx = \frac{dt}{3}\]
\[\text{ Putting x}^3 + 6 x^2 + 5 = t \text{ and }\left( x^2 + 4x \right) dx = \frac{dt}{3}\]
\[ \therefore I = \frac{1}{3}\int\frac{dt}{t}\]
\[ = \frac{1}{3} \text{ ln } \left| t \right| + C\]
\[ = \frac{1}{3}\text{ ln }\left| x^3 + 6 x^2 + 5 \right| + C\]
APPEARS IN
RELATED QUESTIONS
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate: \[\int 2^x \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
