Question

Answer this:

Spamming accounts will be reported. ​

Answer 1

Topic :-

Indefinite Integration

To Integrate :-

$\displaystyle \int \dfrac{1+x^2\ln x}{x+x^2\ln x}\:dx$

Solution :-

$\displaystyle \int \dfrac{1+x^2\ln x}{x+x^2\ln x}\:dx$

Add and Subtract x(1 + lnx) in numerator,

$\displaystyle \int \dfrac{1+x^2\ln x+x(1+\ln x)-x(1+\ln x)}{x(1+x\ln x)}\:dx$

$\displaystyle \int \dfrac{1+x^2\ln x+x+x\ln x-x-x\ln x}{x(1+x\ln x)}\:dx$

Rearranging and grouping terms,

$\displaystyle \int \dfrac{x+x^2\ln x+1+x\ln x-x-x\ln x}{x(1+x\ln x)}\:dx$

$\displaystyle \int \dfrac{x(1+x\ln x)+(1+x\ln x)-x(1+\ln x)}{x(1+x\ln x)}\:dx$

Breaking and Integrating,

$\displaystyle \int \dfrac{x(1+x\ln x)}{x(1+x\ln x)}\:dx+\int \dfrac{(1+x\ln x)}{x(1+x\ln x)}\:dx-\int\dfrac{x(1+\ln x)}{x(1+x\ln x)}\:dx$

$\displaystyle \int dx+\int \dfrac{1}{x}\:dx-\int\dfrac{1+\ln x}{1+x\ln x}\:dx$

$\displaystyle x+\int \dfrac{1}{x}\:dx-\int\dfrac{1+\ln x}{1+x\ln x}\:dx$

$\left(\because \displaystyle \int dz=z+C,where\:C\:is\:constant\:of\:integration\right)$

$\displaystyle x+\ln x-\int\dfrac{1+\ln x}{1+x\ln x}\:dx$

$\left(\because \displaystyle \int \dfrac{1}{z}\:dz=\ln z+C,where\:C\:is\:constant\:of\:integration\right)$

$\bold{Substitute\: \:t=1+x\ln x,}$

$dt=\left(\ln x+\not{x}\cdot\dfrac{1}{\not{x}}\right)dx=(1+\ln x)\:dx$

$\displaystyle x+\ln x-\int\dfrac{1}{t}\:dt$

$x+\ln x-\ln t+C$

$\left(\because \displaystyle \int \dfrac{1}{z}\:dz=\ln z+C,where\:C\:is\:constant\:of\:integration\right)$

Put back value of 't',

$x+\ln x-\ln (1+x\ln x)+C$

$\ln e^x+\ln x-\ln (1+x\ln x)+\ln C'$

$(\because x=\ln e^x)$

$(C=\ln C', C'\:is\:a\:new\:constant)$

$\ln\dfrac{C'xe^x}{1+x\ln x}$

$\left(\because \ln m+\ln n-\ln p-\ln q = \ln \dfrac{mn}{pq} \right)$

Answer :-

$\underline{\boxed{\displaystyle \int \dfrac{1+x^2\ln x}{x+x^2\ln x}\:dx=x+\ln x-\ln (1+x\ln x)+C}}$

$or$

$\underline{\boxed{\displaystyle \int \dfrac{1+x^2\ln x}{x+x^2\ln x}\:dx=\ln\dfrac{C'xe^x}{1+x\ln x}}}$