1. Calculate the generalized integral ∫ [0, + ∞] DX / (100 + x ^ 2). 2. Find the monotone interval of function f (x) = x ^ 2-lnx ^ 2

1. Calculate the generalized integral ∫ [0, + ∞] DX / (100 + x ^ 2). 2. Find the monotone interval of function f (x) = x ^ 2-lnx ^ 2

First triangular substitution
You let x = 10 Tana
So the integral is changed to ∫ (0 to π / 2) 10 (SecA) divided by 100 (SecA) Da = ∫ (0 to π / 2) 1 / 10da = π / 20
The second way: are you ln (x2) or (LNX) side?

Please calculate the generalized integral: ∫ a ^ x ^ 2 DX The interval is from 0 to positive infinity, Where 0

First partial integral ∫ a ^ x x ^ 2 DX = (1 / LNA) ∫ x ^ 2 Da ^ x = a ^ x x ^ 2 / LNA - (1 / LNA) ∫ a ^ x 2x DX = a ^ x ^ 2 / LNA - (1 / LNA) ^ 2 ∫ 2xda ^ x = a ^ x ^ 2 / LNA - (1 / LNA) ^ 2 [2xa ^ X - ∫ 2A ^ xdx] = a ^ x ^ 2 / LNA - (1 / LNA) ^ 2 [2xa ^ X-2 (1 / LNA) a ^ x]

Judgment of convergence and divergence of generalized integral ∫ (upper 1, lower 0) DX / x ^ q! 1. ∫ (up 1 down 0) DX / x ^ q is x = 0. Why? 2. The answer is Q > 1 Q q>1 q

The first floor is wrong. Does the defect have nothing to do with whether there is a definition, but whether the function near it is bounded
When Q0, 1 / x ^ Q has no solution in any neighborhood of 0, so it is a defective integral
Discussing the convergence and divergence of generalized integral is actually discussing whether the limit of the original function exists at the defect point
That is, whether the upper 1 and lower Y 1 / x ^ QDX of LIM (y tends to zero from the positive) integral exist
(the symbol can't be input. Make do with it)
Using Newton Leibniz formula, the above limit is
LIM (y tends to zero from positive) (1-y ^ (1-Q)) / (1-Q) (at Q1)
LIM (y tends to zero from positive direction) (- LNY) (when q = 1)
It is clear that when q = 1 does not exist, it is equivalent to the convergence and divergence of defective integral

Judge the convergence and divergence of the following generalized integrals ∫ x ^ 3E ^ (- x ^ 2) DX, [0, ∞]

Calculate directly
=1/2∫(0,+∞) x^2e^(-x^2)dx^2
=1/2∫(0,+∞) te^(-t)dt
=1/2∫(0,+∞) e^(-t)dt
=1/2

Discrimination of convergence and divergence of generalized integral ∫ (upper e, lower 1 / E) ln|x-1| / (x-1) DX

This integral should be convergent;
∫{x=1/e→e} [ln|x-1|/(x-1)]dx
=∫{x=1/e→1- δ} [ln(1-x) /(x-1)] dx + ∫{x=1- δ →e} [ln(x-1) /(x-1)] dx …… δ →0
=(1/2)ln ² (1-x)|{1/e,1- δ}+ (1/2)ln ² (x-1)|{1+ δ, e}…… δ →0
=ln ²δ- (1/2)ln ² [1-(1/e)]+(1/2)ln ² (e-1)-ln ²δ……δ →0
=(1/2)ln ² {(e-1)/[(e-1)/e]}=1/2

∫ [- ∞, + ∞] (x / √ (1 + X Λ 2)) DX, judge the convergence. If it converges, calculate the value of the generalized integral

Convergence, the generalized integral value is 0, without calculation, just use symmetry, because the integrand function is an odd function, and the upper and lower limits of the integral are symmetrical about the origin. According to the definition of definite integral, the area under the positive half axis curve of X axis is always equal to the area under the negative half axis curve of X axis, and the sign is opposite, so the sum of the two is always 0
Please accept, thank you!