∫ex1+ex2+exdx is equal to

Let,I=∫ex1+ex2+exdxput,ex=t⇒exdx=dt∴ I=∫ex(1+t)(2+t)dtex=∫1(1+t)(2+t)dt Let 1(1+t)(2+t)=A(1+t)+B(2+t)⇒ 1=A(2+t)+B(1+t)=(2A+B)+t(A+B)On equating the coefficients of t and constant term onboth sides, we get A+B=0 and 2A+B=1 On solving both equations, we get A=1 and B=−1∴ I=∫1(1+t)dt−∫1(2+t)dt=log⁡|1+t|−log⁡|2+t|+C=log⁡1+t2+t+C=log⁡1+ex2+ex+C