Integral Calculus

Step 1: Multiply and divide by the constant $\cos(a - b)$, noting $(x - a) - (x - b) = a - b$.
Step 2: Expand numerator: $\cos(x - a)\cos(x - b) + \sin(x - a)\sin(x - b)$.
Step 3: Divide termwise by $\sin(x - a)\cos(x - b)$: integrand = $\cot(x - a) + \tan(x - b)$.
Step 4: Integrate: $\ln|\sin(x - a)| - \ln|\cos(x - b)|$.
Step 5: I = $\frac{1}{\cos(a - b)}[\ln|\sin(x - a)| - \ln|\cos(x - b)|] + C$.

Step 1: Identify the first function $u = x$ and second function $v = e^x$ using the ILATE rule.
Step 2: Apply integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Step 3: Compute $\int e^x dx = e^x$ and $\frac{dx}{dx} = 1$, giving $x e^x - \int 1 \cdot e^x dx$.
Step 4: Evaluate the remaining integral $\int e^x dx = e^x$ to obtain $x e^x - e^x + C$.

Step 1: Express the integrand as a product of $\tan^{-1} x$ and $1$.
Step 2: Choose $u = \tan^{-1} x$ and $v = 1$.
Step 3: Apply integration by parts to obtain $x \tan^{-1} x - \int \frac{x}{1 + x^2} dx$.
Step 4: Rewrite the remaining integral as $\frac{1}{2} \int \frac{2x}{1 + x^2} dx$, which integrates to $\frac{1}{2} \ln(1 + x^2)$.
Step 5: Combine terms to get $x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) + C$.

Step 1: Identify $u = \ln(1 + x)$ and $v = x$.
Step 2: Differentiate $u$ to get $\frac{1}{1 + x}$ and integrate $v$ to get $\frac{x^2}{2}$.
Step 3: Apply integration by parts to get $\frac{x^2}{2} \ln(1 + x) - \frac{1}{2} \int \frac{x^2}{1 + x} dx$.
Step 4: Rewrite $\frac{x^2}{1 + x}$ as $\frac{x^2 - 1 + 1}{x + 1} = (x - 1) + \frac{1}{x + 1}$.
Step 5: Integrate terms individually to obtain $\frac{x^2}{2} \ln(1 + x) - \frac{1}{2} \left[ \frac{x^2}{2} - x + \ln\vert{}1 + x\vert{} \right] + C$.

Step 1: Write the integrand as $\ln(x + \sqrt{x^2 + a^2}) \cdot 1$.
Step 2: Choose $u = \ln(x + \sqrt{x^2 + a^2})$ and $v = 1$.
Step 3: Differentiate $u$ using the chain rule to get $\frac{1}{\sqrt{x^2 + a^2}}$.
Step 4: Apply integration by parts formula: $x \ln(x + \sqrt{x^2 + a^2}) - \int \frac{x}{\sqrt{x^2 + a^2}} dx$.
Step 5: Substitute $x^2 + a^2 = z^2$ so that $x dx = z dz$, which integrates to $\sqrt{x^2 + a^2}$.
Step 6: Combine terms to get $x \ln(x + \sqrt{x^2 + a^2}) - \sqrt{x^2 + a^2} + C$.

Step 1: Choose $u = x^2$ and $dv = e^{-x} dx$.
Step 2: Differentiate $u$ to get $du = 2x \, dx$ and integrate $dv$ to get $v = -e^{-x}$.
Step 3: Apply integration by parts to obtain $-x^2 e^{-x} + 2 \int x e^{-x} \, dx$.
Step 4: Apply integration by parts again to $\int x e^{-x} \, dx$ by taking $u = x$ and $dv = e^{-x} dx$.
Step 5: Integrate to obtain $x(-e^{-x}) - \int (-e^{-x}) \, dx = -x e^{-x} - e^{-x}$.
Step 6: Combine all terms and factor out $-e^{-x}$ to get $-e^{-x} (x^2 + 2x + 2) + C$.

Step 1: Assign $u = x^3$ and $dv = e^{2x} dx$. Then $du = 3x^2 dx$ and $v = \frac{e^{2x}}{2}$.
Step 2: Apply integration by parts to get $I = \frac{1}{2} x^3 e^{2x} - \frac{3}{2} \int x^2 e^{2x} \, dx$.
Step 3: Apply integration by parts to $\int x^2 e^{2x} \, dx$ with $u = x^2$ and $dv = e^{2x} dx$ to get $\frac{1}{2} x^2 e^{2x} - \int x e^{2x} \, dx$.
Step 4: Substitute back to obtain $I = \frac{1}{2} x^3 e^{2x} - \frac{3}{4} x^2 e^{2x} + \frac{3}{2} \int x e^{2x} \, dx$.
Step 5: Apply integration by parts to $\int x e^{2x} \, dx$ to get $\frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}$.
Step 6: Combine all terms together and factor out $\frac{1}{8} e^{2x}$ to obtain $\frac{1}{8} e^{2x} (4x^3 - 6x^2 + 6x - 3) + C$.