Question

$\int \frac{6x+7}{\left(x-5\right)\left(x-4\right)}dx$

Answer 1

Consider the given integral.

$I=\int \frac{6x+7}{(x-5)(x-4)}dx$

$\int \frac{6x+7}{(x-5)(x-4)}dx=\int (\frac{A}{x-5}+\frac{B}{x-4})dx$

$6x+7=A(x-4)+B(x-5)$

Put

$x-4=0$

$x=4$

So,

$6\times 4+7=A\times 0+B(4-5)$

$24+7=-B$

$B=-31$

Put}

$x-5=0$

$x=5$

So,

$6\times 5+7=A(5-4)+B\times 0$

$30+7=A$

$A=37$

Therefore,

$I=\int (\frac{37}{x-5}-\frac{31}{x-4})dx$

$I=37\ln (x-5)-31\ln (x-4)+C$

Hence, the value is $37\ln (x-5)-31\ln (x-4)+C$.