Evaluate: ∫(x²+1)/((x²+4)(x²+25))dx

I=∫(x²+1)/((x²+4)(x²+25))dx

We can use partial fraction decomposition to solve this integral. We assume that:

(x²+1)/((x²+4)(x²+25)) = (Ax+B)/(x²+4) + (Cx+D)/(x²+25)

Multiplying both sides by (x²+4)(x²+25), we get:

x²+1 = (Ax+B)(x²+25) + (Cx+D)(x²+4)

x²+1 = Ax³ + 25Ax + Bx² + 25B + Cx³ + 4Cx + Dx² + 4D

x²+1 = (A+C)x³ + (B+D)x² + (25A+4C)x + (25B+4D)

By comparing coefficients, we obtain the following system of equations:

A + C = 0
B + D = 1
25A + 4C = 0
25B + 4D = 1

From the first and third equations, we get A = 0 and C = 0.
From the second and fourth equations, we get:

B + D = 1
25B + 4D = 1

Solving this system, we find B = -19/21 and D = 40/21

Therefore,

(x²+1)/((x²+4)(x²+25)) = (-19/21)/(x²+4) + (40/21)/(x²+25)

Now we can integrate term by term:

I = ∫((-19/21)/(x²+4) + (40/21)/(x²+25))dx

I = (-19/21)∫(1/(x²+4))dx + (40/21)∫(1/(x²+25))dx

We know that ∫(1/(x²+a²))dx = (1/a)arctan(x/a) + C

So,

I = (-19/21)(1/2)arctan(x/2) + (40/21)(1/5)arctan(x/5) + C

I = (-19/42)arctan(x/2) + (8/21)arctan(x/5) + C