Solved on Feb 27, 2024

Evaluate the integral $\int \frac{9}{8+9 x} d x$ for $x \neq -\frac{8}{9}$ . Type the exact answer.

STEP 1

Assumptions
1. We are given the integral to evaluate: $\int \frac{9}{8+9x} \, dx$
2. The variable of integration is $x$ .
3. $x \neq -\frac{8}{9}$ to avoid division by zero in the integrand.

STEP 2

Identify the form of the integral. The given integral is a rational function where the degree of the numerator is less than the degree of the denominator. This suggests that we can use a $u$ -substitution to simplify the integral.

STEP 3

Let $u$ be the denominator of the fraction in the integrand.
$u = 8 + 9x$

STEP 4

Differentiate $u$ with respect to $x$ to find $du$ .
$\frac{du}{dx} = 9$

STEP 5

Solve for $dx$ in terms of $du$ .
$dx = \frac{du}{9}$

STEP 6

Substitute $8 + 9x$ with $u$ and $dx$ with $\frac{du}{9}$ in the integral.
$\int \frac{9}{8+9x} \, dx = \int \frac{9}{u} \cdot \frac{du}{9}$

STEP 7

Simplify the integral by canceling out the 9's.
$\int \frac{9}{u} \cdot \frac{du}{9} = \int \frac{1}{u} \, du$

STEP 8

The integral of $\frac{1}{u}$ with respect to $u$ is the natural logarithm of the absolute value of $u$ .
$\int \frac{1}{u} \, du = \ln |u| + C$

STEP 9

Substitute back the expression for $u$ to get the integral in terms of $x$ .
$\ln |u| + C = \ln |8 + 9x| + C$

STEP 10

Write the final answer, including the constant of integration $C$ .
$\int \frac{9}{8+9x} \, dx = \ln |8 + 9x| + C$
To check the result, we differentiate the antiderivative.

STEP 11

Differentiate the antiderivative with respect to $x$ .
$\frac{d}{dx}(\ln |8 + 9x| + C)$

STEP 12

Apply the chain rule for differentiation.
$\frac{d}{dx}(\ln |8 + 9x|) = \frac{1}{8 + 9x} \cdot \frac{d}{dx}(8 + 9x)$

STEP 13

Differentiate $8 + 9x$ with respect to $x$ .
$\frac{d}{dx}(8 + 9x) = 9$

STEP 14

Combine the results from steps 12 and 13.
$\frac{1}{8 + 9x} \cdot 9 = \frac{9}{8 + 9x}$

STEP 15

Since the derivative of the antiderivative is the original integrand, our solution is verified.
$\frac{d}{dx}(\ln |8 + 9x| + C) = \frac{9}{8 + 9x}$
Thus, the evaluated integral is correct.
$\int \frac{9}{8+9x} \, dx = \ln |8 + 9x| + C$

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Evaluate the integral $\int \frac{9}{8+9 x} d x$ for $x \neq -\frac{8}{9}$ . Type the exact answer.

STEP 1

Assumptions
1. We are given the integral to evaluate: $\int \frac{9}{8+9x} \, dx$
2. The variable of integration is $x$ .
3. $x \neq -\frac{8}{9}$ to avoid division by zero in the integrand.

STEP 2

Identify the form of the integral. The given integral is a rational function where the degree of the numerator is less than the degree of the denominator. This suggests that we can use a $u$ -substitution to simplify the integral.

STEP 3

Let $u$ be the denominator of the fraction in the integrand.
$u = 8 + 9x$

STEP 4

Differentiate $u$ with respect to $x$ to find $du$ .
$\frac{du}{dx} = 9$

STEP 5

Solve for $dx$ in terms of $du$ .
$dx = \frac{du}{9}$

STEP 6

Substitute $8 + 9x$ with $u$ and $dx$ with $\frac{du}{9}$ in the integral.
$\int \frac{9}{8+9x} \, dx = \int \frac{9}{u} \cdot \frac{du}{9}$

STEP 7

Simplify the integral by canceling out the 9's.
$\int \frac{9}{u} \cdot \frac{du}{9} = \int \frac{1}{u} \, du$

STEP 8

The integral of $\frac{1}{u}$ with respect to $u$ is the natural logarithm of the absolute value of $u$ .
$\int \frac{1}{u} \, du = \ln |u| + C$

STEP 9

Substitute back the expression for $u$ to get the integral in terms of $x$ .
$\ln |u| + C = \ln |8 + 9x| + C$

STEP 10

Write the final answer, including the constant of integration $C$ .
$\int \frac{9}{8+9x} \, dx = \ln |8 + 9x| + C$
To check the result, we differentiate the antiderivative.

STEP 11

Differentiate the antiderivative with respect to $x$ .
$\frac{d}{dx}(\ln |8 + 9x| + C)$

STEP 12

Apply the chain rule for differentiation.
$\frac{d}{dx}(\ln |8 + 9x|) = \frac{1}{8 + 9x} \cdot \frac{d}{dx}(8 + 9x)$

STEP 13

Differentiate $8 + 9x$ with respect to $x$ .
$\frac{d}{dx}(8 + 9x) = 9$

STEP 14

Combine the results from steps 12 and 13.
$\frac{1}{8 + 9x} \cdot 9 = \frac{9}{8 + 9x}$

STEP 15

Since the derivative of the antiderivative is the original integrand, our solution is verified.
$\frac{d}{dx}(\ln |8 + 9x| + C) = \frac{9}{8 + 9x}$
Thus, the evaluated integral is correct.
$\int \frac{9}{8+9x} \, dx = \ln |8 + 9x| + C$

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Evaluate the integral ∫98+9xdx\int \frac{9}{8+9 x} d x∫8+9x9​dx for x≠−89x \neq -\frac{8}{9}x=−98​. Type the exact answer.

STEP 1

Assumptions1. We are given the integral to evaluate: ∫98+9x dx\int \frac{9}{8+9x} \, dx∫8+9x9​dx2. The variable of integration is xxx.3. x≠−89x \neq -\frac{8}{9}x=−98​ to avoid division by zero in the integrand.

STEP 2

Identify the form of the integral. The given integral is a rational function where the degree of the numerator is less than the degree of the denominator. This suggests that we can use a uuu-substitution to simplify the integral.

STEP 3

Let uuu be the denominator of the fraction in the integrand.u=8+9xu = 8 + 9xu=8+9x

STEP 4

Differentiate uuu with respect to xxx to find dududu.dudx=9\frac{du}{dx} = 9dxdu​=9

STEP 5

Solve for dxdxdx in terms of dududu.dx=du9dx = \frac{du}{9}dx=9du​

STEP 6

Substitute 8+9x8 + 9x8+9x with uuu and dxdxdx with du9\frac{du}{9}9du​ in the integral.∫98+9x dx=∫9u⋅du9\int \frac{9}{8+9x} \, dx = \int \frac{9}{u} \cdot \frac{du}{9}∫8+9x9​dx=∫u9​⋅9du​

STEP 7

Simplify the integral by canceling out the 9's.∫9u⋅du9=∫1u du\int \frac{9}{u} \cdot \frac{du}{9} = \int \frac{1}{u} \, du∫u9​⋅9du​=∫u1​du

STEP 8

The integral of 1u\frac{1}{u}u1​ with respect to uuu is the natural logarithm of the absolute value of uuu.∫1u du=ln⁡∣u∣+C\int \frac{1}{u} \, du = \ln |u| + C∫u1​du=ln∣u∣+C

STEP 9

Substitute back the expression for uuu to get the integral in terms of xxx.ln⁡∣u∣+C=ln⁡∣8+9x∣+C\ln |u| + C = \ln |8 + 9x| + Cln∣u∣+C=ln∣8+9x∣+C

STEP 10

Write the final answer, including the constant of integration CCC.∫98+9x dx=ln⁡∣8+9x∣+C\int \frac{9}{8+9x} \, dx = \ln |8 + 9x| + C∫8+9x9​dx=ln∣8+9x∣+CTo check the result, we differentiate the antiderivative.

STEP 11

Differentiate the antiderivative with respect to xxx.ddx(ln⁡∣8+9x∣+C)\frac{d}{dx}(\ln |8 + 9x| + C)dxd​(ln∣8+9x∣+C)

STEP 12

Apply the chain rule for differentiation.ddx(ln⁡∣8+9x∣)=18+9x⋅ddx(8+9x)\frac{d}{dx}(\ln |8 + 9x|) = \frac{1}{8 + 9x} \cdot \frac{d}{dx}(8 + 9x)dxd​(ln∣8+9x∣)=8+9x1​⋅dxd​(8+9x)

STEP 13

Differentiate 8+9x8 + 9x8+9x with respect to xxx.ddx(8+9x)=9\frac{d}{dx}(8 + 9x) = 9dxd​(8+9x)=9

STEP 14

Combine the results from steps 12 and 13.18+9x⋅9=98+9x\frac{1}{8 + 9x} \cdot 9 = \frac{9}{8 + 9x}8+9x1​⋅9=8+9x9​

STEP 15

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Evaluate the integral $\int \frac{9}{8+9 x} d x$ for $x \neq -\frac{8}{9}$ . Type the exact answer.

Assumptions
1. We are given the integral to evaluate: $\int \frac{9}{8+9x} \, dx$
2. The variable of integration is $x$ .
3. $x \neq -\frac{8}{9}$ to avoid division by zero in the integrand.

Identify the form of the integral. The given integral is a rational function where the degree of the numerator is less than the degree of the denominator. This suggests that we can use a $u$ -substitution to simplify the integral.

Let $u$ be the denominator of the fraction in the integrand.
$u = 8 + 9x$

Differentiate $u$ with respect to $x$ to find $du$ .
$\frac{du}{dx} = 9$

Solve for $dx$ in terms of $du$ .
$dx = \frac{du}{9}$

Substitute $8 + 9x$ with $u$ and $dx$ with $\frac{du}{9}$ in the integral.
$\int \frac{9}{8+9x} \, dx = \int \frac{9}{u} \cdot \frac{du}{9}$

Simplify the integral by canceling out the 9's.
$\int \frac{9}{u} \cdot \frac{du}{9} = \int \frac{1}{u} \, du$

The integral of $\frac{1}{u}$ with respect to $u$ is the natural logarithm of the absolute value of $u$ .
$\int \frac{1}{u} \, du = \ln |u| + C$

Substitute back the expression for $u$ to get the integral in terms of $x$ .
$\ln |u| + C = \ln |8 + 9x| + C$

Write the final answer, including the constant of integration $C$ .
$\int \frac{9}{8+9x} \, dx = \ln |8 + 9x| + C$
To check the result, we differentiate the antiderivative.

Differentiate the antiderivative with respect to $x$ .
$\frac{d}{dx}(\ln |8 + 9x| + C)$

Apply the chain rule for differentiation.
$\frac{d}{dx}(\ln |8 + 9x|) = \frac{1}{8 + 9x} \cdot \frac{d}{dx}(8 + 9x)$

Differentiate $8 + 9x$ with respect to $x$ .
$\frac{d}{dx}(8 + 9x) = 9$

Combine the results from steps 12 and 13.
$\frac{1}{8 + 9x} \cdot 9 = \frac{9}{8 + 9x}$