Math  /  Calculus

QuestionEvaluate. e7x(e7x+7)4dx=\int \frac{e^{7 x}}{\left(e^{7 x}+7\right)^{4}} d x= \square

Studdy Solution

STEP 1

1. The integral is of the form f(x)(f(x))ndx\int \frac{f'(x)}{(f(x))^n} \, dx, which suggests a substitution method.
2. We will use substitution to simplify the integral.

STEP 2

1. Choose an appropriate substitution.
2. Differentiate the substitution to find dudu.
3. Rewrite the integral in terms of uu.
4. Integrate with respect to uu.
5. Substitute back to express the result in terms of xx.

STEP 3

Choose the substitution u=e7x+7 u = e^{7x} + 7 .

STEP 4

Differentiate the substitution to find du du :
ddx(e7x+7)=7e7x \frac{d}{dx}(e^{7x} + 7) = 7e^{7x}
So, du=7e7xdx du = 7e^{7x} \, dx .
To express dx dx in terms of du du , we have:
dx=du7e7x dx = \frac{du}{7e^{7x}}

STEP 5

Rewrite the integral in terms of u u :
Since u=e7x+7 u = e^{7x} + 7 , we have e7x=u7 e^{7x} = u - 7 .
The integral becomes:
e7x(e7x+7)4dx=u7u417e7xdu \int \frac{e^{7x}}{(e^{7x} + 7)^4} \, dx = \int \frac{u - 7}{u^4} \cdot \frac{1}{7e^{7x}} \, du
This simplifies to:
17u3du \int \frac{1}{7u^3} \, du

STEP 6

Integrate with respect to u u :
17u3du=17u3du \int \frac{1}{7u^3} \, du = \frac{1}{7} \int u^{-3} \, du
The integral of u3 u^{-3} is:
17u22=114u2 \frac{1}{7} \cdot \frac{u^{-2}}{-2} = -\frac{1}{14} u^{-2}

STEP 7

Substitute back to express the result in terms of x x :
Since u=e7x+7 u = e^{7x} + 7 , we have:
114u2=114(e7x+7)2 -\frac{1}{14} u^{-2} = -\frac{1}{14} (e^{7x} + 7)^{-2}
Thus, the evaluated integral is:
114(e7x+7)2 \boxed{-\frac{1}{14(e^{7x} + 7)^2}}

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