Integrate following integrals.

1.integral ax+b/(sqrt(ax^2+2bx+c)dx
2.integral 1+x/(1+x^2)dx
3.integral e^x+1/e^x dx

(1+x) dx/(1+x^2)

= dx/(1+x^2) + x dx/(1+x^2)
tan^-1 ( x) + .5 ln(1+x^2) + constant

3.integral e^x+1/e^x dx

I guess maybe you mean'
3.integral (e^x+1) /e^x dx
which would be'
dx + e^-x dx
x -e^-x + constant

Judging by the level of problem of the 1st and 3rd, I think the second is as it was typed

find ∫ 1+x/(1+x^2)dx
= x + (1/2) ln(1+x^2) + c

To integrate these functions, we will use different techniques based on the structure of the integrals.

1. Integral of (ax+b) / √(ax^2 + 2bx + c) dx:
This integral involves a rational function divided by a square root. To solve it, we can make a substitution. Let's substitute u = ax^2 + 2bx + c. Then, du = (2ax + 2b) dx.

Now, we need to replace dx and the numerator. We can divide both numerator and denominator by a^2 to simplify the expression:

∫ (ax + b) / √(ax^2 + 2bx + c) dx
= ∫ [(ax + b) / (a√(x^2 + (2b/a)x + c/a^2))] dx
= (1/a) ∫ [(ax + b) / √(x^2 + (2b/a)x + c/a^2)] dx

With the substitution, we have:
= (1/a) ∫ (1/2√(c/a^2)) * (2ax + 2b) / √(u) du
= (1/a) (1/√(c/a^2)) ∫ (ax + b) / √(u) du
= (2/√ac) ∫ √u du
= (2/√ac) * (2/3) u^(3/2) + C

Finally, substituting back u = ax^2 + 2bx + c, we have:
= (4/3√ac) * (ax^2 + 2bx + c)^(3/2) + C

2. Integral of (1 + x) / (1 + x^2) dx:
This integral involves a rational function. We can split it into two parts using partial fraction decomposition.

First, decompose (1 + x) / (1 + x^2) into partial fractions:
(1 + x) / (1 + x^2) = A(1 + x) + B(1 - x^2) / (1 + x^2)

Multiply both sides by (1 + x^2):
1 + x = A(1 + x)(1 + x^2) + B(1 - x^2)

To determine the values of A and B, we can compare the coefficients. Expanding the equation, we get:
1 + x = A + Ax^2 + Ax + Ax^3 + B - Bx^2

Comparing the coefficients of each power of x:
1st-degree term: 1 + x = Ax + Bx
Constant term: 1 = A + B

From the first equation, A = 0 and B = 1. Substituting these values, the integral becomes:
∫ (1 + x) / (1 + x^2) dx = ∫ (1 - x^2) / (1 + x^2) dx
= ∫ dx - ∫ (x^2) / (1 + x^2) dx
= x - arctan(x) + C

3. Integral of e^x + 1 / e^x dx:
To integrate this function, we can use the linearity property of integration. Split the integral into two parts:

∫ (e^x + 1/e^x) dx
= ∫ e^x dx + ∫ 1/e^x dx
= e^x - ∫ e^(-x) dx

Now, integrate the second term:
∫ e^(-x) dx = -e^(-x) + C

Combining the results, we have:
∫ (e^x + 1/e^x) dx
= e^x - (-e^(-x)) + C
= e^x + e^(-x) + C

let z = a x^2 + 2 b x + c

then dz/dx = 2 a x + 2 b = 2 (ax+b)
so
dx = dz/2(ax+b)
then
(ax+b) dx/(sqrt(ax^2+2bx+c) = (dz/2)/z^.5
=(1/2)(z^-.5 dz)
integrate to get
= (.5)(z^.5) / .5 = z^.5
so
sqrt (a x^2 + 2 b x + c) + constant
check arithmetic I did it fast