   Chapter 8.4, Problem 69E

Chapter
Section
Textbook Problem

PUTNAM EXAM CHALLENGEEvaluate ∫ 0 1 ln ( x + 1 ) x 2 + 1 d x .

To determine

To calculate: The value of thefollowing definite integral I=01ln(x+1)x2+1dx.

Explanation

Given:

The given definite integral I=01ln(x+1)x2+1dx.

Formula used:

1x2+1dx=tan1x+c

Calculation:

Consider the definite integral;

I=01ln(x+1)x2+1dx

Now, substitute x=1u1+u so, the term x2+1 can be expressed as;

x=1u1+ux+1=21+ux2+1=2+2u2(1+u)2

The differentiate x=1u1+u to get;

dx=2(1+u)2du

The limit of integration is to be change as the variable changed.

Nowsubstitutevariable x with 0 the value of u will be,

x=1u1+u0=1u1+uu=1

When substituting variable x with 1 the value of u will be;

1=1u1+u1+u=1uu=0

Thus, the definite integration can be shown as;

I=10

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