In some of the cases we can split the integrand into the sum of the two functions such that the integration of one of them by parts produces an integral which cancels the other integral. Suppose we have an integral of the type∫[f(x)h(x)+g(x)]dxLet ∫f(x)h(x)dx=I1 and ∫g(x)dx=I2Integrating I1 by parts, we getI1=f(x)∫h(x)dx−∫{f′(x)∫h(x)dx}dxNow find ∫(1log x−1(log x)2)dx (x > 0)

In some of the cases we can split the integrand into the sum of the tw

1.	In some of the cases we can split the integrand into the sum of the two functions such that the integration of one of them by parts produces an integral which cancels the other integral. Suppose we have an integral of the type∫[f(x)h(x)+g(x)]dxLet ∫f(x)h(x)dx=I1 and ∫g(x)dx=I2Integrating I1 by parts, we getI1=f(x)∫h(x)dx−∫{f′(x)∫h(x)dx}dxNow find ∫(1log x−1(log x)2)dx (x > 0)
Answer» In some of the cases we can split the integrand into the sum of the two functions such that the integration of one of them by parts produces an integral which cancels the other integral. Suppose we have an integral of the type $\int [f (x) h (x) + g (x)] d x$ Let $\int f (x) h (x) d x = I_{1}$ and $\int g (x) d x = I_{2}$ Integrating $I_{1}$ by parts, we get $I_{1} = f (x) \int h (x) d x - \int {f^{'} (x) \int h (x) d x} d x$ Now find $\int (\frac{1}{l o g x} - \frac{1}{(l o g x)^{2}}) d x$ (x > 0)