Integration by parts of In(x+1)/√x+1
Let I = int {ln(x+1)/√x + 1} dx
Let I1 = int {ln(x+1)/√x} dx
Integrating by parts,
I1 = 2√x.ln(x+1) – int{2√x/(x+1)} dx
I1 = 2√x.ln(x+1) – 2.I2
where I2 = int{√x/(x+1)} dx.
Let u = √x
Then du = 1/(2√x) dx
Or, dx = 2u du
Substituting for the above into I2,
I2 = int{u/(u^2+1)} 2u.du
I2 = 2.int{u^2/(u^2+1)}.du = 2.int{(u^2+1 – 1)/(u^2+1)}.du = 2.int{1 – 1/(u^2+1)}.du
I2 = 2u – 2.arctan(u)
Hence, I1 = 2√x.ln(x+1) – 2.I2 = 2√x.ln(x+1) – 4u + 4.arctan(u)
I1 = 2√x.ln(x+1) – 4√x + 4.arctan(√x)
Hence, I = I1 + int{1} dx
I = 2√x.ln(x+1) – 4√x + 4.arctan(√x) + x