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

Question

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

UpStudy AI · Accepted Answer

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.

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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.
The integrals are: 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

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

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