please solve this problem in detail, with explanation---The image contains a mathematical expression, specifically a definite integral.
**Mathematical Expression:**
The integral is from 0 to +infinity.
The integrand is a fraction with numerator ln x and denominator 1 + x^2.
The differential is dx.
The complete expression is:
$\int_{0}^{+\infty} \frac{\ln x}{1 + x^2} dx$
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We need to evaluate the definite integral from 0 to positive infinity of ln x divided by 1 plus x squared, dx. This is a challenging integral that requires a clever substitution technique. Let's visualize the integrand function first.
To solve this integral, we'll use the substitution x equals 1 over u. This gives us dx equals negative 1 over u squared du. When x approaches 0 from the positive side, u approaches positive infinity. When x approaches positive infinity, u approaches 0 from the positive side. This substitution will help us discover a key property of this integral.
Now we substitute everything into the integral. First, we replace x with 1 over u, and dx with negative 1 over u squared du. The logarithm of 1 over u becomes negative ln u. The denominator 1 plus 1 over u squared becomes u squared plus 1, all over u squared. After simplifying the algebra, we get the integral from positive infinity to 0 positive of ln u over u squared plus 1, du.
Here comes the key insight. When we reverse the limits of integration from positive infinity to 0 positive, to 0 positive to positive infinity, we introduce a negative sign. Since u is just a dummy variable, we can replace it with x. This gives us I equals negative I. Adding I to both sides gives us 2I equals 0, which means I equals 0. The integral evaluates to zero!
In conclusion, the definite integral from 0 to positive infinity of ln x over 1 plus x squared dx equals zero. This remarkable result occurs because the positive and negative areas under the curve exactly cancel each other out. The substitution method revealed this hidden symmetry, showing that the integral is equal to its own negative, which can only happen when the value is zero. This elegant solution demonstrates the power of clever substitution techniques in evaluating challenging integrals.