Evaluate [tex] \int \frac{48 x+30}{8 x^2+10 x+4} d x [/tex] where [tex] 8 x^2+10 x+4\ \textgreater \ 0 [/tex]

First substitute [tex]u =[/tex] $\square$
Then [tex]\int \frac{48 x+30}{8 x^2+10 x+4} d x=\int[/tex] $\square$ du Now integrate with respect to u to get $\square$ [tex]+C[/tex] So [tex]\int \frac{48 x+30}{8 x^2+10 x+4} d x=[/tex] $\square$ [tex]+ C[/tex]

To evaluate the integral ∫ 8 x 2 + 10 x + 4 48 x + 30 ​ d x , we use the substitution u = 8 x 2 + 10 x + 4 . This leads us to the integral 3 ln ∣ u ∣ + C , which simplifies to 3 ln ∣8 x 2 + 10 x + 4∣ + C after substituting back for u .
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Answered by Anonymous | 2025-07-04

Substitute u = 8 x 2 + 10 x + 4 , so d u = ( 16 x + 10 ) d x .
Rewrite the integral as ∫ 8 x 2 + 10 x + 4 48 x + 30 ​ d x = ∫ 8 x 2 + 10 x + 4 3 ( 16 x + 10 ) ​ d x = ∫ u 3 ​ d u .
Integrate to get 3 ln ∣ u ∣ + C .
Substitute back u = 8 x 2 + 10 x + 4 to obtain the final answer: 3 ln ∣8 x 2 + 10 x + 4∣ + C ​ .

Explanation

Problem Analysis We are asked to evaluate the integral ∫ 8 x 2 + 10 x + 4 48 x + 30 ​ d x where 0"> 8 x 2 + 10 x + 4 > 0 . This looks like a good candidate for u-substitution.

Finding du Let's set u = 8 x 2 + 10 x + 4 . Then, we need to find d u . Taking the derivative of u with respect to x , we get d x d u ​ = 16 x + 10 . Thus, d u = ( 16 x + 10 ) d x .

Rewriting the Integral Now, we want to rewrite the numerator 48 x + 30 in terms of 16 x + 10 . Notice that 48 x + 30 = 3 ( 16 x + 10 ) . So, we can rewrite the integral as ∫ 8 x 2 + 10 x + 4 48 x + 30 ​ d x = ∫ 8 x 2 + 10 x + 4 3 ( 16 x + 10 ) ​ d x

Substituting u and du Using our substitution u = 8 x 2 + 10 x + 4 and d u = ( 16 x + 10 ) d x , we can rewrite the integral in terms of u :
∫ 8 x 2 + 10 x + 4 3 ( 16 x + 10 ) ​ d x = ∫ u 3 ​ d u

Integrating with respect to u Now, we integrate with respect to u :
∫ u 3 ​ d u = 3 ∫ u 1 ​ d u = 3 ln ∣ u ∣ + C

Substituting back for x Finally, we substitute back u = 8 x 2 + 10 x + 4 to get the integral in terms of x :
3 ln ∣ u ∣ + C = 3 ln ∣8 x 2 + 10 x + 4∣ + C

Final Answer Therefore, the integral is: ∫ 8 x 2 + 10 x + 4 48 x + 30 ​ d x = 3 ln ∣8 x 2 + 10 x + 4∣ + C So, the answers to the blanks are: First substitute u = 8 x 2 + 10 x + 4 Then ∫ 8 x 2 + 10 x + 4 48 x + 30 ​ d x = ∫ 3 d u Now integrate with respect to u to get 3 ln ∣ u ∣ + C So $\int \frac{48 x+30}{8 x^2+10 x+4} d x= 3 \ln |8x^2 + 10x + 4| + C


Examples
Imagine you're calculating the flow rate of a fluid through a pipe where the velocity profile is described by a rational function. Integrating such a function, similar to the one in this problem, helps determine the total volume of fluid passing through a cross-section of the pipe over time. This is crucial in chemical engineering for designing and optimizing processes involving fluid transport.

Answered by GinnyAnswer | 2025-07-04