Evaluate : (i) `inte^(x)((1)/(x)-(1)/(x^(2)))dx` (ii) `inte^(x)((1)/(x^(2))-(2)/(x^(3)))dx`
(iii) `inte^(x){sin^(-1)x+(1)/(sqrt(1-x^(2)))}dx` (iv) `inte^(x)(tanx+logsecx)dx`

Explanation

We have
(i) `I=inte^(x){(1)/(x)+(-(1)/(x^(2)))}dx`
`=inte^(x){f(x)+f'(x)}dx,"where"f(x)=(1)/(x)andf'(x)=(-1)/(x^(2))`
`=e^(x)*f(x)+c=e^(x)*(1)/(x)+C=(e^(x))/(x)+C`.
(ii) `I=inte^(x){(1)/(x^(2))+((-2)/(x^(3)))}dx`
`=inte^(x){f(x)+f'(x)}dx,"where"f(x)=(1)/(x^(2))andf'(x)=(-2)/(x^(3))`
`=e^(x)*f(x)+C=e^(x)*(1)/(x^(2))+C=(e^(x))/(x^(2))+C`.
(iii) `I=inte^(x){sin^(-1)x+(1)/(sqrt(1-x^(2)))}dx`
`inte^(x){f(x)+f'(x)}dx,"where"f(x)=sin^(-1)xandf'(x)=(1)/(sqrt(1-x^(2)))`
`=e^(x)*f(x)+C=e^(x)*sin^(-1)x+C=e^(x)sin^(-1)x+C`.
(iv) `I=inte^(x)(tanx+logsecx)dx`
`=inte^(x){f(x)+f'(x)}dx,"wheref(x)=log(secx)andf'(x)=(1)/(secx)*secxtanx=tanx`
`=e^(x)f(x)+C=e^(x)log(secx)+C`.