Question

integral 2x+3/(x+3)(x^+4)dx,notequal to-3​

Answer 1

Answer:

Solution



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∫x2(1−x21)dx

=∫(x2−1)dx

=2+1x2+1−x+C

=3x3−x+C

where C is constant

Step-by-step explanation:

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Answer 2

Appropriate Question

Evaluate the following integral

$\rm :\longmapsto\:\displaystyle\int\sf \frac{2x + 3}{(x + 3)( {x}^{2} + 4)}dx$

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \frac{2x + 3}{(x + 3)( {x}^{2} + 4)}dx$

To evaluate this integral, we use the Method of Partial Fractions.

So, Let assume that

$\rm :\longmapsto\:\dfrac{2x + 3}{(x + 3)( {x}^{2} + 4)} = \dfrac{a}{x + 3} + \dfrac{bx + c}{ {x}^{2} + 4} - - - (1)$

$\rm :\longmapsto\:2x + 3 = a({x}^{2} + 4) + (bx + c)(x + 3) - - - (2)$

On substituting x = - 3, we get

$\rm :\longmapsto\: - 6 + 3 = a({( - 3)}^{2} + 4) + (bx + c)( - 3 + 3)$

$\rm :\longmapsto\: - 3 = a(9 + 4)$

$\rm\implies \: \: \boxed{\tt{ \: \: a \: = \: - \: \frac{3}{13} \: \: }}$

On substituting x = 0, in equation (2), we get

$\rm :\longmapsto\:0 + 3 = a(0 + 4) + (0 + c)(0 + 3)$

$\rm :\longmapsto\:3 = 4a + 3c$

$\rm :\longmapsto\:3 = - 4 \times\dfrac{3}{13} + 3c$

$\rm :\longmapsto\:3 = - \dfrac{12}{13} + 3c$

$\rm :\longmapsto\:3 + \dfrac{12}{13} = 3c$

$\rm :\longmapsto\: \dfrac{39 + 12}{13} = 3c$

$\rm :\longmapsto\: \dfrac{51}{13} = 3c$

$\rm\implies \:\boxed{\tt{ \: \: c \: = \: \frac{17}{13} \: \: }}$

On substituting x = 1, in equation (2), we get

$\rm :\longmapsto\:2 + 3 = a(1 + 4) + (b + c)(1+ 3)$

$\rm :\longmapsto\:5 = 5a + 4b + 4c$

$\rm :\longmapsto\:5 = - \dfrac{15}{13} + 4b + \dfrac{68}{13}$

$\rm :\longmapsto\:4b = 5 + \dfrac{15}{13} - \dfrac{68}{13}$

$\rm :\longmapsto\:4b = \dfrac{65 + 15 - 68}{13}$

$\rm :\longmapsto\:4b = \dfrac{80 - 68}{13}$

$\rm :\longmapsto\:4b = \dfrac{12}{13}$

$\rm\implies \:\boxed{\tt{ \: \: b \: = \: \frac{3}{13} \: \: }}$

So, on substituting the values of a, b and c, in equation (1), we get

$\rm :\longmapsto\:\dfrac{2x + 3}{(x + 3)( {x}^{2} + 4)} = \dfrac{ - 3}{13(x + 3)} + \dfrac{3x + 17}{13( {x}^{2} + 4)}$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{2x + 3}{(x + 3)( {x}^{2} + 4)}dx =\displaystyle\int\sf \dfrac{ - 3}{13(x + 3)}dx+\displaystyle\int\sf \dfrac{3x + 17}{13( {x}^{2} + 4)}dx$

$\rm \:  =  \: - \dfrac{1}{3}\displaystyle\int\sf \frac{1}{x + 3}dx + \dfrac{3}{13}\displaystyle\int\sf \frac{x}{ {x}^{2} + 4}dx + \dfrac{17}{13}\displaystyle\int\sf \frac{dx}{ {x}^{2} + 4}$

can be further rewritten as

$\rm \:  =  \: - \dfrac{1}{3}\displaystyle\int\sf \frac{1}{x + 3}dx + \dfrac{3}{26}\displaystyle\int\sf \frac{2x}{ {x}^{2} + 4}dx + \dfrac{17}{13}\displaystyle\int\sf \frac{dx}{ {x}^{2} + 4}$

We know,

$\boxed{\tt{ \displaystyle\int\sf \frac{1}{x} \: dx \: = \: log |x| + c}}$

$\boxed{\tt{ \displaystyle\int\sf \frac{1}{ {x}^{2} + {a}^{2} } \: dx \: = \: \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c}}$

$\boxed{\tt{ \displaystyle\int\sf \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c}}$

So, using this result, we get

$\rm \:  =  \: - \dfrac{3}{13} log |x + 3| + \dfrac{3}{26}log | {x}^{2} + 4| + \dfrac{17}{13} \times \dfrac{1}{2} {tan}^{ - 1} \dfrac{x}{2} + c'$

$\rm \:  =  \: - \dfrac{3}{13} log |x + 3| + \dfrac{3}{26}log | {x}^{2} + 4| + \dfrac{17}{26} {tan}^{ - 1} \dfrac{x}{2} + c'$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$