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**

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