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
Evaluate : `int_0^3dx/(9+x^2)`
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "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))`
