[latex]\int (5x^4 + 2x) e^{x^5 + x^2}dx[/latex]

[latex]\int \frac{2x}{x^2 - 5}dx[/latex]

1. [latex]\int e^{-3x}dx[/latex].
   
   [latex]\frac{1}{-3} e^{-3x} + C[/latex]

   ans
2. [latex]\int \left( \frac{1}{2}x - 1 \right)^4dx[/latex]
   
   [latex].4 \left( \frac{1}{2} x - 1 \right)^5 + C[/latex]

   ans

   .

3. [latex]\int_0^2 (5x - 3)^3dx[/latex].
   
   [latex]116[/latex]

   ans
4. [latex]\int \frac{1}{7x - 2}dx[/latex]
   
   [latex]\frac{1}{7} \ln(7x - 2) + C[/latex]

   ans
5. [latex]\int \frac{1}{\sqrt{0.5x - 1}}dx[/latex]
   
   [latex]4 \sqrt{0.5x - 1} + C[/latex]

   ans

1. [latex]\int (8x^3) (x^4 + 1)^2dx[/latex]
   
   1. We see that [latex]u = x^4 + 1[/latex] in this case.
   2. We see [latex]\frac{du}{dx} = 4x^3[/latex], so [latex]dx = \frac{1}{4x^3} du[/latex].
   3. We have
      \begin{align*}
      & = \int (8x^3) u^2 \frac{1}{4x^3} du \\
      & = \int 8 u^2 \frac{1}{4}du \\
      & = 2 \int u^2du
      \end{align*}
   4. Integrating we have

      [latex]2 \int u^2du = 2 \left( \frac{1}{3} u^3 \right ) + C = \frac{2}{3} u^3 + C[/latex]

   5. Substitution of [latex]u = x^4 + 1[/latex] yields

      [latex]\frac{2}{3} (x^4 + 1)^3 + C[/latex]

   ans
2. [latex]\int (x^4 + 1)^2 (8x^3)dx[/latex]
   
   This is exactly the same as the previous problem, just written a different way. No need to redo work.

   ans
3. [latex]\int (3x^2 - 1) e^{x^3 - x}dx[/latex]
   
   1. We see that [latex]u = x^3 - x[/latex] in this case.
   2. We see [latex]\frac{du}{dx} = 3x^2 - 1[/latex], so [latex]dx = \frac{1}{3x^2 - 1} du[/latex].
   3. We have
      \begin{align*}
      & = \int (3x^2 – 1) e^u \frac{1}{3x^2 – 1} du \\
      & = \int e^udu
      \end{align*}
   4. Integrating we have

      [latex]\int e^udu = e^u + C[/latex]

   5. Substitution of [latex]u = x^3 - x[/latex] yields

      [latex]e^{x^3 - x} + C[/latex]

   ans
4. [latex]\int (x^2 - \frac{1}{3}) e^{x^3 - x}dx[/latex]
   
   [latex]\frac{1}{3} e^{x^3 - x} + C[/latex]

   ans
5. [latex]\int (e^x+ x)^5 (e^x+1)dx[/latex]
   
   [latex]\frac{1}{6} (e^x + x)^6 + C[/latex]

   ans
6. [latex]\int \frac{2x}{x^2 - 1}dx[/latex]
   
   [latex]\ln(x^2 -1) + C[/latex]

   ans
7. [latex]\int e^x \sqrt{e^x + 1}dx[/latex]
   
   [latex]\frac{2}{3}(e^x + 1)^{3/2} + C[/latex]

   ans
8. [latex]\int \frac{\sin(x)}{\cos(x)}dx[/latex]
   
   [latex]-\ln(\cos(x)) + C[/latex]

   ans
9. [latex]\int \frac{\sqrt{ \ln(x) }}{x}dx[/latex]
   
   [latex]\frac{2}{3} (\ln(x))^{3/2} + C[/latex]

   ans
10. [latex]\int \frac{x}{(x^2 - 5)^3}dx[/latex]
    
    1. We see that [latex]u = x^2 - 5[/latex] in this case.
    2. We see [latex]\frac{du}{dx} = 2x[/latex], so [latex]dx = \frac{1}{2x} du[/latex].
    3. We have
       \begin{align*}
       & = \int \frac{x}{u^3}\frac{1}{2x}du \\
       & = \int \frac{1}{2u^3}du \\
       & = \int \frac{1}{2}u^{-3}du
       \end{align*}
    4. Integrating we have

       [latex]\int \frac{1}{2}u^{-3}du = - \frac{1}{4} u^{-2} + C[/latex]

    5. Substitution of [latex]u = x^2 - 5[/latex] yields

       [latex]- \frac{1}{4} (x^2 - 5)^{-2} + C[/latex]

    ans
11. [latex]\int \frac{\cos(x)}{\sqrt{\sin(x) + 1}}dx[/latex]
    
    [latex]2\sqrt{\sin(x) + 1} + C[/latex]

    ans
12. [latex]\int \frac{1}{x (2 \ln(x) + 1)^4}dx[/latex].
    
    [latex]-\frac{1}{6(2 \ln(x) + 1)^3} + C[/latex]

    ans
13. [latex]\int (e^{5x} + 1)^9 (e^{5x} )dx[/latex].
    
    [latex]\frac{1}{2} (e^{5x} + 1)^{10} + C[/latex]

    ans