X'3+4x+1/ ( x-1)(x+1)(x+3) dx 2. 6x+7/(x+2)'2 dx ?

1. To integrate X^3 + 4x + 1 / (x-1)(x+1)(x+3) dx, we first need to perform partial fraction decomposition on the given expression.

X^3 + 4x + 1 / (x-1)(x+1)(x+3) = A/(x-1) + B/(x+1) + C/(x+3)

Multiplying both sides by (x-1)(x+1)(x+3), we get:

X^3 + 4x + 1 = A(x+1)(x+3) + B(x-1)(x+3) + C(x-1)(x+1)

Expanding and equating coefficients, we get:

X^3 + 4x + 1 = A(x^2 + 4x + 3) + B(x^2 + 2x - 3) + C(x^2 - x - 1)

Now, solve for A, B, and C by comparing coefficients. Once you have the values of A, B, and C, you can integrate each term separately.

2. To integrate 6x + 7 / (x+2)^2 dx, you can rewrite the expression as:

6x + 7 / (x+2)^2 = 6x/(x+2)^2 + 7/(x+2)^2

Now, you can integrate each term separately.

For the first term, 6x/(x+2)^2, you can use u-substitution where u = x+2:

Let u = x+2
Then, du = dx

The integral becomes:

∫ 6x/(x+2)^2 dx = ∫ 6(u-2)/u^2 du
= ∫ (6u/u^2 - 12/u^2) du
= ∫ (6/u - 12/u^2) du
= 6ln|u| + 12/u + C
= 6ln|x+2| + 12/(x+2) + C

For the second term, 7/(x+2)^2, you can use a simple substitution:

Let t = x+2
Then, dt = dx

The integral becomes:

∫ 7/(x+2)^2 dx = ∫ 7/t^2 dt
= -7/t + C
= -7/(x+2) + C

Therefore, the final integral of 6x + 7 / (x+2)^2 dx is:

6ln|x+2| + 12/(x+2) - 7/(x+2) + C
= 6ln|x+2| + 5/(x+2) + C

1.

$$\int \frac{3x+1}{(x-1)(x+1)(x-3)} dx$$

Partial Fraction Decomposition:

$$\frac{3x+1}{(x-1)(x+1)(x-3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x-3}$$

Solving for the constants:

$$A = 1$$
$$B = 1$$
$$C = -1$$

Therefore,

$$\frac{3x+1}{(x-1)(x+1)(x-3)} = \frac{1}{x-1} + \frac{1}{x+1} - \frac{1}{x-3}$$

Now, we can integrate each term:

$$\int \frac{1}{x-1} dx = \ln|x-1| + C_1$$
$$\int \frac{1}{x+1} dx = \ln|x+1| + C_2$$
$$\int \frac{1}{x-3} dx = \ln|x-3| + C_3$$

Combining the constants, we get the final answer:

$$\int \frac{3x+1}{(x-1)(x+1)(x-3)} dx = \ln|x-1| + \ln|x+1| - \ln|x-3| + C$$

2.

$$\int \frac{6x^7}{(x^2+2)^2} dx$$

u-Substitution:

Let \(u = x^2+2\). Then \(du = 2x dx\).

Substituting:

$$\frac{6x^7}{(x^2+2)^2} dx = \frac{6}{2} \cdot \frac{u^7}{u^2} du = 3u^5 du$$

Now, we can integrate:

$$\int 3u^5 du = \frac{3}{6} u^6 + C = \frac{1}{2} u^6 + C$$

Substituting back \(u = x^2+2\):

$$\int \frac{6x^7}{(x^2+2)^2} dx = \frac{1}{2} (x^2+2)^6 + C$$