Intriguing integrals! - one

Let us tackle this using REVERSE INTEGRATION BY PARTS we will start with the integral $I_1 =\int_0^a \frac{1}{1+x} dx$ we know the value of this integral {it is ln(a+1) } ..... We will continuously apply the so called wrong way of integration by parts by taking u as 1/1+x and v as 1 ..... The formula for integration by parts is given by $\int uv dx = u \times \int v dx - \int { \int v dx} \frac{du}{dx} dx$ now we will continuously apply this until we get our required integral

first step yeilds $I_1 = \frac{x}{1+x}|_0^a+ \int_0^a \frac{x}{(1+x)^2}$ $I_1 = \frac{a}{1+a} + \frac{x^2}{2\times(1+x)^2} |_0^a + \int_0^a \frac{x^2}{(1+x)^3} dx$ so we now get an idea of where we are going ... the next step immediately gives $I_1 = \frac{a}{1+a} + \frac{a^2}{2\times(1+a)^2} + \frac{a^3}{3\times(1+a)^3} + \int_0^a \frac{x^3}{(1+x)^4} dx$ we will plug in a=3 and shift the constant terms to the left to get the answer .... $ln(3+1) - \frac{a}{1+a} - \frac{a^2}{2\times(1+a)^2} -\frac{a^3}{3\times(1+a)^3} = \int_0^3 \frac{x^3}{(1+x)^4} dx$ $\int_0^3 \frac{x^3}{(1+x)^4} dx = ln(4) - \frac{75}{64}$

hence the answer is 143.

NOTE - this integral simply is the infinite series for natural logarithm with the omission of few initial terms (decided by the power of x in numerator in the integral ) Hence picking up the whole ln function and subtracting a few terms we get the value of the integral