Question

integrate (3x + 4)/(x ^ 2 + 6x + 5) dx​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\int\rm \frac{3x + 4}{ {x}^{2} + 6x + 5} \: dx$

Let assume that

$\rm :\longmapsto\:I = \displaystyle\int\rm \frac{3x + 4}{ {x}^{2} + 6x + 5} \: dx$

can be rewritten as

$\rm \: = \:\displaystyle\int\rm \frac{3x + 4}{ {x}^{2} + 5x + x + 5 } \: dx$

$\rm \: = \:\displaystyle\int\rm \frac{3x + 4}{ x(x + 5) + 1(x + 5) } \: dx$

$\rm \: = \:\displaystyle\int\rm \frac{3x + 4}{(x + 5)(x + 1)} \: dx$

Now using Partial Fraction, we have

$\rm :\longmapsto\:\dfrac{3x + 4}{(x + 5)(x + 1)} = \dfrac{a}{(x + 5)} + \dfrac{b}{(x + 1)} - - (1)$

On taking LCM, we get

$\rm :\longmapsto\:3x + 4 = a(x + 1) + b(x + 5)$

On substituting x = - 5, we get

$\rm :\longmapsto\:3( - 5) + 4 = a( - 5 + 1) + b( - 5 + 5)$

$\rm :\longmapsto\: - 15 + 4 = - 4 a$

$\rm :\longmapsto\: - 11 = - 4 a$

$\rm \implies\:\boxed{ \tt{ \: a = \frac{11}{4} \: }}$

On substituting x = - 1, we get

$\rm :\longmapsto\:3( - 1) + 4 = a( - 1 + 1) + b( - 1 + 5)$

$\rm :\longmapsto\: - 3 + 4 = 4b$

$\rm :\longmapsto\: 1= 4b$

$\rm \implies\:\boxed{ \tt{ \: b \: = \: \frac{1}{4} \: }}$

On substituting the values of a and b, in equation (1), we get

$\rm :\longmapsto\:\dfrac{3x + 4}{(x + 5)(x + 1)} = \dfrac{11}{4(x + 5)} + \dfrac{1}{4(x + 1)}$

On integrating both sides w. r. t. x, we get

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{3x + 4}{(x + 5)(x + 1)}dx =\displaystyle\int\rm \dfrac{11}{4(x + 5)}dx + \displaystyle\int\rm \dfrac{1}{4(x + 1)}dx$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{1}{x} \: dx = logx + c \: }}$

So, using this, we get

$\rm :\longmapsto\:\displaystyle\int\rm \frac{3x + 4}{(x + 5)(x + 1)}dx = \frac{11}{4}log |x + 5| + \frac{1}{4}log |x + 1| + c$

Hence,

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{3x + 4}{ {x}^{2} + 6x + 5}dx = \frac{11}{4}log |x + 5| + \frac{1}{4}log |x + 1| + c \: }}$

More to know :-

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

Answer 2

Answer:

$\\ </p><p>\boxed{ \tt{ \: \displaystyle\int\rm \frac{3x + 4}{ {x}^{2} + 6x + 5}dx = \frac{11}{4}log |x + 5| + \frac{1}{4}log |x + 1| + c \: }} \\ \\ </p><p></p><p>$

Step-by-step explanation:

Hope it helps you