Mathematics, 12.12.2020 16:40 krojas015

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

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Answers

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

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

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

Step-by-step 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

[Integrand] Rewrite: $\displaystyle \int {\bigg( 6e^\big{4x} \cdot \frac{e^\big{arctan(3e^{4x})}}{1 + 9e^\big{8x}} \bigg)} \, dx = \int {\frac{6e^\big{arctan(3e^{4x}) + 4x}}{1 + 9e^\big{8x}}} \, dx$ [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.

Set u: $\displaystyle u = 4x$ [u] Differentiate [Basic Power Rule, Multiplied Constant]: $\displaystyle du = 4 \ dx$

Step 4: Integrate Pt. 3

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

Step 5: Integrate Pt. 4

Set variables for u-substitution #2.

Set v: $\displaystyle v = 9e^{2u} + 1$ [v] Differentiate [Exponential Differentiation, Chain Rule]: $\displaystyle dv = 18e^{2u} \ du$ [v] U-Solve: $\displaystyle u = ln \Big( \frac{\sqrt{v - 1}}{3} \Big)$ [dv] U-Solve: $\displaystyle du = \frac{e^{-2u}}{18} \ dv$ [U-Solve] Rewrite u: $\displaystyle e^u = \frac{\sqrt{v - 1}}{3}$

Step 6: Integrate Pt. 5

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

Step 7: Integrate Pt. 6

Set variables for u-substitution #3.

Set z: $\displaystyle z = arctan(\sqrt{v - 1})$ [z] Differentiate [Arctrig Differentiation, Chain Rule]: $\displaystyle dz = \frac{1}{2v\sqrt{v - 1}} \ dv$

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