\( \begin{array}{ll}\int x^{2} e^{3 x^{3}} d x & \text { c. } \int\left(1+\frac{1}{x^{2}}\right) e^{x-\frac{1}{x}} d x \\ \int \sec ^{2} x e^{\tan x+1} d x & \text { f. } \int e^{\sin x \cos x} \cos 2 x d x \\ \int \frac{e^{2 x}}{e^{x}+1} d x\end{array} \)

We wish to evaluate the following integrals:  1. ∫ x² e^(3x³) dx  2. ∫ (1 + (1/x²)) e^(x – 1/x) dx  3. ∫ sec²x e^(tan x + 1) dx  4. ∫ e^(sin x cos x) cos 2x dx  5. ∫ e^(2x)/(e^x + 1) dx Below are step‐by–step solutions for each. ────────────────────────────── 1. ∫ x² e^(3x³) dx • Notice that since the exponent is 3x³, it is natural to let   u = 3x³. • Then, differentiate:   du/dx = 9x² ⟹ du = 9x² dx. • Solve for x² dx:   x² dx = du/9. • Substitute into the integral:   ∫ x² e^(3x³) dx = ∫ e^u (du/9) = (1/9) ∫ e^u du. • Integrate:   (1/9) e^u + C = (1/9) e^(3x³) + C. ────────────────────────────── 2. ∫ [1 + (1/x²)] e^(x – 1/x) dx • Let   u = x – 1/x. • Differentiate:   du/dx = 1 + 1/x² ⟹ du = (1 + 1/x²) dx. • This shows that the differential (1 + 1/x²) dx is exactly du. Then we have:   ∫ [1 + (1/x²)] e^(x – 1/x) dx = ∫ e^u du. • Integrate:   e^u + C = e^(x–1/x) + C. ────────────────────────────── 3. ∫ sec²x e^(tan x + 1) dx • Let   u = tan x. • Differentiate:   du/dx = sec²x, so du = sec²x dx. • Then the integral becomes:   ∫ e^(u + 1) du = e^1 ∫ e^u du = e ∫ e^u du. • Integrate:   e e^u + C = e e^(tan x) + C. • We can also write this as:   e^(tan x + 1) + C. ────────────────────────────── 4. ∫ e^(sin x cos x) cos 2x dx • Note that the derivative of sin x cos x is:   d/dx (sin x cos x) = cos^2 x – sin^2 x = cos 2x. • Thus, let   u = sin x cos x ⟹ du = cos 2x dx. • Then the integral becomes:   ∫ e^u du. • Integrate:   e^u + C = e^(sin x cos x) + C. ────────────────────────────── 5. ∫ e^(2x)/(e^x + 1) dx • Notice that e^(2x) = (e^x)². Let   u = e^x. • Differentiate:   du/dx = e^x ⟹ dx = du/u. • Substitute into the integral:   ∫ u²/(u + 1) · (du/u) = ∫ u/(u + 1) du. • Perform algebra: Write u/(u + 1) as:   u/(u + 1) = 1 – 1/(u + 1). • Thus, the integral becomes:   ∫ [1 – 1/(u + 1)] du = ∫ 1 du – ∫ 1/(u + 1) du. • Integrate:   = u – ln|u + 1| + C. • Return to x by recalling u = e^x:   = e^x – ln(e^x + 1) + C. ────────────────────────────── Final Answers 1. ∫ x² e^(3x³) dx = (1/9) e^(3x³) + C 2. ∫ [1 + (1/x²)] e^(x – 1/x) dx = e^(x – 1/x) + C 3. ∫ sec²x e^(tan x + 1) dx = e^(tan x + 1) + C 4. ∫ e^(sin x cos x) cos 2x dx = e^(sin x cos x) + C 5. ∫ e^(2x)/(e^x + 1) dx = e^x – ln(e^x + 1) + C Any answer that is algebraically equivalent is correct.