MHB Brad's questions at Yahoo Answers regarding finding anti-derivatives

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The discussion focuses on finding the antiderivatives of two functions: -8(4x-1)^(1/2) and -2(3x^4-5)^2. For the first function, a substitution method is used, leading to the result of -4/3(4x-1)^(3/2) + C. The second function is expanded and integrated term by term, resulting in -2(x^9 - 6x^5 + 25x) + C. Detailed steps for both integrations are provided to aid understanding.
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Here are the questions:

Find the antiderivative?


Struggling with these two antiderivative:

-8(4x-1)^1/2

-2(3x^4-5)^2

Please explain how you worked them out with full working

I have posted a link there to this topic so the OP can see my work.
 
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Re: Brad's questions at Yahoo Answers regarding finding anti-derivatives

Hello Brad,

1.) $$I=\int-8(4x-1)^{\frac{1}{2}}\,dx$$

Let's use the substitution:

$$u=4x-1\,\therefore\,du=4\,dx$$

And we may now state:

$$I=-2\int u^{\frac{1}{2}}\,du$$

Using the power rule for integration, we may write:

$$I=-2\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right)+C=-\frac{4}{3}u^{\frac{3}{2}}+C$$

Back-substituting for $u$, there results:

$$I=-\frac{4}{3}(4x-1)^{\frac{3}{2}}+C$$

2.) $$I=\int -2(3x^4-5)^2\,dx$$

Bringing the constant factor in the integrand out front and expanding the binomial, we have:

$$I=-2\int 9x^8-30x^4+25\,dx$$

Applying the power rule term by term, we find:

$$I=-2\left(9\frac{x^9}{9}-30\frac{x^5}{5}+25x \right)+C$$

$$I=-2\left(x^9-6x^5+25x \right)+C$$
 
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