Question

Integration of the following

Answer 1

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

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{3x + 4}{6x + 7}dx$

On multiply and divide by 2, we get

$\rm \: \: = \: \dfrac{1}{2} \displaystyle\int\tt \dfrac{6x + 8}{6x + 7}dx$

$\rm \: \: = \: \dfrac{1}{2} \displaystyle\int\tt \dfrac{6x + 7 + 1}{6x + 7}dx$

$\rm \: \: = \: \dfrac{1}{2} \displaystyle\int\tt \dfrac{6x + 7}{6x + 7}dx + \dfrac{1}{2} \displaystyle\int\tt \dfrac{1}{6x + 7}dx$

$\rm \: \: = \: \dfrac{1}{2} \displaystyle\int\tt 1dx + \dfrac{1}{2} \displaystyle\int\tt \dfrac{1}{6x + 7}dx$

$\rm \: \: = \: \dfrac{1}{2}x + \dfrac{1}{12}log |6x + 7| + c$

$\red{ \boxed{ \because \sf{ \:\displaystyle\int\tt {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c}}}$

$\red{ \boxed{ \because \sf{ \:\displaystyle\int\tt \frac{1 }{x}dx = log(x) + c }}}$

Hence,

$\bf\implies \: \displaystyle\int\tt \dfrac{3x + 4}{6x + 7}dx = \: \dfrac{1}{2}x + \dfrac{1}{12}log |6x + 7| + c$

Aliter Method

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{3x + 4}{6x + 7}dx$

We use substitution method,

$\red{\rm :\longmapsto\:Put \: 6x + 7 = y}$

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

$\red{\rm :\longmapsto\ \: 6dx = dy}$

Also,

$\rm :\longmapsto\:x = \dfrac{y - 7}{6}$

Hence,

Given integral reduced to

$\rm \: \: = \: \displaystyle\int\tt \dfrac{3\bigg(\dfrac{y - 7}{6}\bigg) + 4}{y} \times \dfrac{1}{6} dy$

$\rm \: \: = \: \dfrac{1}{6} \displaystyle\int\tt \dfrac{\bigg(\dfrac{y - 7}{2}\bigg) + 4}{y} \: dy$

$\rm \: \: = \:\dfrac{1}{6} \displaystyle\int\tt \dfrac{\bigg(\dfrac{y - 7 + 8}{2}\bigg)}{y} \: dy$

$\rm \: \: = \:\dfrac{1}{6} \displaystyle\int\tt \dfrac{\bigg(\dfrac{y + 1}{2}\bigg)}{y} \: dy$

$\rm \: \: = \:\dfrac{1}{12} \displaystyle\int\tt \dfrac{y + 1}{y}dy$

$\rm \: \: = \:\dfrac{1}{12} \displaystyle\int\tt (1 \: + \: \dfrac{1}{y})dy$

$\rm \: \: = \: \dfrac{1}{12}(y + log |y|) + c$

$\rm \: \: = \: \dfrac{1}{12}(6x + 7 + log |6x + 7|) + c$

Additional Information :-

$\rm :\longmapsto\:\displaystyle\int\tt {e}^{x}dx = {e}^{x} + c$

$\rm :\longmapsto\:\displaystyle\int\tt cosx \: dx = sinx \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\tt sinx \: dx = - \: cosx \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\tt tanx \: dx = \: log(secx) \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\tt cotx \: dx = \: log(sinx) \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\tt cosecx \: dx = \: log(cosecx - cotx) \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\tt secx \: dx = \: log(secx + tanx) \: + \: c$

Answer 2

We have,

$\displaystyle \int \dfrac{3x + 4}{6x + 7}\ dx$

$\displaystyle = \dfrac12 \int \dfrac{6x + 8}{6x + 7}\ dx$

$\displaystyle = \dfrac12 \int \dfrac{6x + 7 + 1}{6x + 7}\ dx$

$\displaystyle = \dfrac12 \int \dfrac{6x + 7}{6x + 7}\ dx +\dfrac12 \int \dfrac1{6x + 7}\ dx$

$\displaystyle = \dfrac12 \int1\ dx +\dfrac1{12} \int \dfrac6{6x + 7}\ dx$

$\displaystyle = \dfrac x2 +\dfrac1{12} \int \dfrac{d(6x + 7)}{6x + 7}$

$\underline{ = \dfrac x2 +\dfrac1{12} \ln|6x + 7| + C}$