Integration Techniques Summary

Approach: Inspect the structure of the integrand: $\frac{x^2-1}{x^2}$ looks complicated, but numerator and denominator have the same degree, so a single step of division turns it into $1 - \frac{1}{x^2}$. This is exactly what the decomposition method is for — splitting a complicated function into simpler pieces that the basic formulas handle, and then integrating the parts separately using linearity.

Steps:

1. Inspect and simplify: $\frac{x^2-1}{x^2} = \frac{x^2}{x^2} - \frac{1}{x^2} = 1 - \frac{1}{x^2} = 1 - x^{-2}$.
2. Split by linearity: $\int \left(1 - x^{-2}\right) dx = \int 1 \, dx - \int x^{-2} dx$.
3. Apply the basic formulas: $\int 1 \, dx = x + C_1$ and $\int x^{-2} dx = \frac{x^{-1}}{-1} + C_2 = -\frac{1}{x} + C_2$.
4. Combine the terms and the constants: $\int \frac{x^2-1}{x^2} dx = x + \frac{1}{x} + C$.
5. Verify by differentiating: $\frac{d}{dx}\left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^2} = \frac{x^2-1}{x^2}$, which matches the original integrand.

Answer: $\int \frac{x^2-1}{x^2} dx = x + \frac{1}{x} + C$.

Approach: The integrand contains $x$ together with $x^2$, and the derivative of $x^2$ is exactly the $2x$ that appears, so put $u = x^2$ and convert the whole differential at once. After integrating, the new variable $u$ must be replaced by the original variable $x$, and the constant of integration must be restored.

Steps:

1. Let $u = x^2$, so $du = 2x \, dx$ and therefore $x \, dx = \frac{1}{2} du$.
2. Substitute: $\int x \cos(x^2) dx = \frac{1}{2} \int \cos u \, du = \frac{1}{2} \sin u + C$.
3. Back-substitute: replacing $u = x^2$ gives $\int x \cos(x^2) dx = \frac{1}{2} \sin(x^2) + C$.
4. Verify by differentiating: $\frac{d}{dx}\left[\frac{1}{2}\sin(x^2)\right] = \frac{1}{2} \cdot 2x \cos(x^2) = x \cos(x^2)$, which matches the original integrand.

Answer: $\int x \cos(x^2) dx = \frac{1}{2} \sin(x^2) + C$.

Approach: The denominator is not in the form of a basic formula, so nothing can be applied directly. Inspecting it, $x^2+2x+2$ can be completed to a square: $x^2+2x+2 = (x+1)^2 + 1$, and $\int \frac{du}{u^2+1} = \arctan u + C$ is a known basic formula. After completing the square, one simple substitution $u = x+1$ finishes the job.

Steps:

1. Complete the square: $x^2 + 2x + 2 = (x+1)^2 + 1$, that is, $x^2+2x+2 = (x+1)^2 + 1^2$.
2. Let $u = x+1$, so $du = dx$ and the integral becomes $\int \frac{du}{u^2+1}$.
3. Apply the basic formula $\int \frac{du}{u^2+1} = \arctan u + C$ to get $\int \frac{dx}{x^2+2x+2} = \arctan u + C$.
4. Back-substitute $u = x+1$: $\int \frac{dx}{x^2+2x+2} = \arctan(x+1) + C$.
5. Verify by differentiating: $\frac{d}{dx}\arctan(x+1) = \frac{1}{1+(x+1)^2} = \frac{1}{x^2+2x+2}$, which matches the original integrand.

Answer: $\int \frac{dx}{x^2+2x+2} = \arctan(x+1) + C$.