Question

The indefinite integral of ((6e^4x)(e^arctan(3e^4x)))/(1+9e^8x)

Answer 1

`\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = \frac{e^\big{arctan(3e^(4x))}}{2} + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Integration

• Integrals
• [Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

U-Substitution

• U-Solve

Explanation:

Step 1: Define

Identify

`\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx`

Step 2: Integrate Pt. 1

1. [Integrand] Rewrite:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = \int {\frac{6e^\big{arctan(3e^(4x)) + 4x}}{1 + 9e^\big{8x}}} \, dx`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = 6\int {\frac{e^\big{arctan(3e^(4x)) + 4x}}{1 + 9e^\big{8x}}} \, dx`

Step 3: integrate Pt. 2

Set variables for u-substitution.

1. Set u:
   `\displaystyle u = 4x`
2. [u] Differentiate [Basic Power Rule, Multiplied Constant]:
   `\displaystyle du = 4 \ dx`

Step 4: Integrate Pt. 3

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = (3)/(2)\int {\frac{4e^\big{arctan(3e^(4x)) + 4x}}{1 + 9e^\big{8x}}} \, dx`
2. [Integral] U-Substitution:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = (3)/(2)\int {\frac{e^\big{arctan(3e^u) + u}}{1 + 9e^\big{2u}}} \, du`

Step 5: Integrate Pt. 4

Set variables for u-substitution #2.

1. Set v:
   `\displaystyle v = 9e^(2u) + 1`
2. [v] Differentiate [Exponential Differentiation, Chain Rule]:
   `\displaystyle dv = 18e^(2u) \ du`
3. [v] U-Solve:
   `\displaystyle u = ln \Big( (√(v - 1))/(3) \Big)`
4. [dv] U-Solve:
   `\displaystyle du = (e^(-2u))/(18) \ dv`
5. [U-Solve] Rewrite u:
   `\displaystyle e^u = (√(v - 1))/(3)`

Step 6: Integrate Pt. 5

1. [Integral] U-Solve:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = (3)/(2)\int {\frac{e^\big{arctan(3((√(v - 1))/(3))) + ln((√(v - 1))/(3))}}{1 + 9((√(v - 1))/(3))^2} (1)/(18e^(2u))\, dv`
2. [Integral] Simplify:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = (3)/(2)\int {\frac{√(v - 1)e^\big{arctan(√(v - 1))}}{3[1 + v - 1]} (1)/(18((√(v - 1))/(3))^2)\, dv`
3. [Integral] Simplify:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = (3)/(2)\int {\frac{√(v - 1)e^\big{arctan(√(v - 1))}}{3v} (1)/(2(v - 1))\, dv`
4. [Integral] Simplify:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = (3)/(2)\int {\frac{e^\big{arctan(√(v - 1))}}{6v√(v - 1)} \, dv`
5. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^(4x))}}{1 + 9e^\big{8x}} \bigg)} \, dx = (1)/(4)\int {\frac{e^\big{arctan(√(v - 1))}}{v√(v - 1)} \, dv`

Step 7: Integrate Pt. 6

Set variables for u-substitution #3.

1. Set z:
   `\displaystyle z = arctan(√(v - 1))`
2. [z] Differentiate [Arctrig Differentiation, Chain Rule]:
   `\displaystyle dz = (1)/(2v√(v - 1)) \ dv`

See attachment for rest of work (would not fit entire answer in answering box).