by&nbsp;using&nbsp;substitution&nbsp;method,&nbsp;integrate∫e4x3e4x-2dx

∫e4x3e4x-2dxLet&nbsp;u2=3e4x-2. Then&nbsp;du2=12e4xdx, so&nbsp;112du2=e4xdx. Rewrite using&nbsp;u2&nbsp;and&nbsp;du2.Let&nbsp;u2=3e4x-2. Find&nbsp;du2dx.Differentiate&nbsp;3e4x-2.ddx[3e4x-2]By the Sum Rule, the derivative of&nbsp;3e4x-2&nbsp;with respect to&nbsp;x&nbsp;is&nbsp;ddx[3e4x]+ddx[-2].ddx[3e4x]+ddx[-2]Evaluate&nbsp;ddx[3e4x].12e4x+ddx[-2]Differentiate using the Constant Rule.12e4xRewrite the problem using&nbsp;u2&nbsp;and&nbsp;du2.∫1u2⋅112du2Simplify.Multiply&nbsp;1u2&nbsp;by&nbsp;112.∫1u2⋅12du2Move&nbsp;12&nbsp;to the left of&nbsp;u2.∫112u2du2Since&nbsp;112&nbsp;is constant with respect to&nbsp;u2, move&nbsp;112&nbsp;out of the integral.112∫1u2du2The integral of&nbsp;1u2&nbsp;with respect to&nbsp;u2&nbsp;is&nbsp;ln(|u2|).112(ln(|u2|)+C)Simplify.112ln(|u2|)+CReplace all occurrences of&nbsp;u2&nbsp;with&nbsp;3e4x-2.112ln(|3e4x−2|)+C

Check the solution to "byusingsubstitutionmethod,integrate∫e4x3e4x-2dx"

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jemijemi181

Answered question

2022-04-09

$b y u \sin g s u b s t i t u t i o n m e t h o d, i n t e g r a t e \int \frac{e^{4 x}}{3 e^{4 x} - 2} d x$

Answer & Explanation

nick1337

Expert2022-06-08Added 777 answers

$\int \frac{e^{4 x}}{3 e^{4 x} - 2} d x$

Let $u_{2} = 3 e^{4 x} - 2$ . Then $d u_{2} = 12 e^{4 x} d x$ , so $\frac{1}{12} d u_{2} = e^{4 x} d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Let $u_{2} = 3 e^{4 x} - 2$ . Find $\frac{d u_{2}}{d x}$ .

Differentiate $3 e^{4 x} - 2$ .

$\frac{d}{d x} [3 e^{4 x} - 2]$

By the Sum Rule, the derivative of $3 e^{4 x} - 2$ with respect to $x$ is $\frac{d}{d x} [3 e^{4 x}] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [3 e^{4 x}] + \frac{d}{d x} [- 2]$

Evaluate $\frac{d}{d x} [3 e^{4 x}]$ .

$12 e^{4 x} + \frac{d}{d x} [- 2]$

Differentiate using the Constant Rule.

$12 e^{4 x}$

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int \frac{1}{u_{2}} \cdot \frac{1}{12} d u_{2}$

Simplify.

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{12}$ .

$\int \frac{1}{u_{2} \cdot 12} d u_{2}$

Move $12$ to the left of $u_{2}$ .

$\int \frac{1}{12 u_{2}} d u_{2}$

Since $\frac{1}{12}$ is constant with respect to $u_{2}$ , move $\frac{1}{12}$ out of the integral.

$\frac{1}{12} \int \frac{1}{u_{2}} d u_{2}$

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{12} (\ln (| u_{2} |) + C)$

Simplify.

$\frac{1}{12} \ln (| u_{2} |) + C$

Replace all occurrences of $u_{2}$ with $3 e^{4 x} - 2$ .

$\frac{1}{12} \ln (| 3 e^{4 x} - 2 |) + C$

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