The value of the integral ∫03 dxx+1+5x+1 is

. The integrand can be written as x+1−5x+1−4xSo the given integral is of the form  I1+I2        I1=−14∫03 x+1xdx=−12∫12 t2t2−1dt=−12∫12 1+1t2−1dt=−121−12limt→1 log⁡|t−1|t+1x+1=t2and     I2=14∫03 5x+1xdx=14∫14 2u2u2−1duu2=5x+1=123+12log⁡35−12limu→1 log⁡|u−1|u+1Thus    I1+I2=1+14log⁡35