Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[I = \int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x}\]
\[ = \int\frac{\left( 5 x^2 + 20x + 6 \right) dx}{x \left( x^2 + 2x + 1 \right)}\]
\[ = \int\frac{\left( 5 x^2 + 20x + 6 \right) dx}{x \left( x + 1 \right)^2}\]
\[\text{Let }\frac{5 x^2 + 20x + 6}{x \left( x + 1 \right)^2} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{\left( x + 1 \right)^2}\]
\[ \Rightarrow \frac{5 x^2 + 20x + 6}{x \left( x + 1 \right)^2} = \frac{A \left( x + 1 \right)^2 + B \left( x \right) \left( x + 1 \right) + C \left( x \right)}{x \left( x + 1 \right)^2}\]
\[ \Rightarrow 5 x^2 + 20x + 6 = A \left( x^2 + 2x + 1 \right) + B \left( x^2 + x \right) + Cx\]
\[ \Rightarrow 5 x^2 + 20x + 6 = \left( A + B \right) x^2 + \left( 2A + B + C \right) x + A\]
\[\text{Equating coefficients of like terms}\]
\[A + B = 5 . . . . . \left( 1 \right)\]
\[2A + B + C = 20 . . . . . \left( 2 \right)\]
\[ A = 6 . . . . . \left( 3 \right)\]
\[\text{Solving (1), (2) and (3), we get}\]
\[A = 6 \])
\[B = - 1\]
\[C = 9\]
\[ \therefore \frac{5 x^2 + 20x + 6}{x \left( x + 1 \right)^2} = \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{\left( x + 1 \right)^2}\]
\[ \Rightarrow I = 6\int\frac{dx}{x} - \int\frac{dx}{x + 1} + 9\int\frac{dx}{\left( x + 1 \right)^2}\]
\[ = 6 \log \left| x \right| - \log \left| x + 1 \right| - \frac{9}{x + 1} + C\]
APPEARS IN
संबंधित प्रश्न
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
If `int(2x^(1/2))/(x^2) dx = k . 2^(1/x) + C`, then k is equal to ______.
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
