Evaluate these integrals...?

This would be a lot of work to do them all, so I'll just do the one I find

interesting and perhaps other contributors will try the remaining questions.

5.) integral(15*(arctan(x) / x^2) dx). I'll just forget about the factor 15 until later.

Use integration by parts. Let u = arctan(x) and dv = (1/x^2) dx.

Then du = (1/(1 + x^2)) dx and v = -(1/x), and so

integral((arctan(x) / x^2) dx) = -(1/x)*arctan(x) + integral((1/(x*(1 + x^2))) dx).

For this last integral, let w = x^2; then dw = 2x dx, and so the result is

integral((1/(x*(1 + x^2))) dx) = (1/2)*integral((1/(w*(1 + w))) dw) =

(1/2)*integral(((1/w) - (1/(w + 1))) dw =

(1/2)*ln lwl - (1/2)*ln lw + 1l + C = (1/2)*ln lx^2l - (1/2)*ln(x^2 + 1) + C,

where I could drop the absolute value signs for (x^2 + 1) as that is always

positive. So multiplying by the factor 15 we end up with a full solution of

15*[ln lxl - (1/2)*ln(x^2 + 1) - (1/x)*arctan(x)] + C,

noting that (1/2)*ln lx^2l = (1/2)*2*ln lxl = ln lxl.

Edit: I'll give #3 and #4 a try now.

3) Forgetting about the factor 26 for now, we proceed with partial fractions:

1/((x - 1)*(x^2 + 25)) = (A/(x - 1)) + ((Bx + C)/(x^2 + 25)) ------>

1 = A*(x^2 + 25) + (Bx + C)*(x - 1) ------>

1 = A*x^2 + 25*A + B*x^2 + Cx - Bx - C ------>

1 = (A + B)*x^2 + (C - B)*x + (25*A - C). Now equate coefficient of like powers

to find that A + B = 0, C - B = 0 and 25*A - C = 1.

So (A + B) + (C - B) = 0 ------> A + C = 0.

Thus (25*A - C) + (A + C) = 1 + 0 ------> 26*A = 1 -----> A = 1/26.

Therefore C = -1/26 = B, and so

integral((26/((x - 1)*(x^2 + 25))) dx) =

26*[(1/26)*integral((1/(x - 1)) dx) - (1/26)*integral(((x + 1)/(x^2 + 25)) dx)] =

ln lx - 1l - integral((x/(x^2 + 25)) dx) - integral((1/(x^2 + 25)) dx).

With integral((x/(x^2 + 25)) dx) we let u = x^2 + 25; then du = 2x dx,

and so we end up with integral((1/2)*(1/u) du) = (1/2)*ln lul = (1/2)*ln(x^2 + 25),

where I could drop the absolute value signs as x^2 + 25 > 0 for all x.

For integral((1/(x^2 + 25)) dx) we let x = 5*tan(z); then dx = 5*sec^2(z) dz and

x^2 + 25 = 25*tan^2(z) + 25 = 25*(tan^2(z) + 1) = 25*sec^2(z).

Our integral then becomes integral((5*sec^2(z) / (25*sec^2(z))) dz) =

(1/5)*integral(dz) = (1/5)*z = (1/5)*arctan(x/5).

Thus our full solution is

ln lx - 1l - (1/2)*ln(x^2 + 25) - (1/5)*arctan(x/5) + C for some constant C.

4.) Forgetting about the factor 7 for now, we then have

1/((x*(x^4 + 8)) = (A/x) + ((B*x^3 + C*x^2 + D*x + E)/(x^4 + 8)) ------->

1 = A*(x^4 + 8) + (B*x^3 + C*x^2 + D*x + E)*x ------>

1 = (A + B)*x^4 + C*x^3 + D*x^2 + E*x + 8A. Equating coefficients of like

powers, we see that A + B = 0, C = D = E = 0 and 8A = 1.

Thus A = 1/8 and B = -1/8, and so

integral((7/((x*(x^4 + 8)) dx) = (7/8)*[integral((1/x) dx) - integral((x^3/(x^4 + 8)) dx)] =

(7/8)*ln lxl - (7/32)*integral((1/u) du) = (7/8)*ln lxl - (7/32)*ln(x^4 + 8) + C

where in the last integral I made the substitution u = x^4 + 8; du = 4*x^3 dx.

Also note that I didn't put absolute value signs around x^4 + 8 because

x^4 + 8 > 0 for all real x.