Linked Questions

7
votes
2answers
8k views

Can an indefinite integral have multiple answers? (Besides the ' + C') [duplicate]

So I came across with this integral today in my midterm: $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2 $$ And I got two correct answers: $$\frac {\sec^2(\pi x)}{4\pi} +C$$ And $$\frac {\tan^2(\pi x)}{...
3
votes
2answers
5k views

Solving indefinite integrals gives multiple answers. Are all those answers correct? [duplicate]

While solving problems on indefinite integrals many a times I get answers which are different from those given in my text book's answer keys page. I then verify my solution steps to ensure that even ...
1
vote
2answers
540 views

Same integration with 2 different answers? [duplicate]

$$\int x(x^2+2)^4\,dx $$ When we do this integration with u substitution we get $$\frac{(x^2+2)^5}{10}$$ as $u=x^2+2$ $du=2x\,dx$ $$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$ ...
1
vote
3answers
2k views

Integration : different answers from two methods [duplicate]

There seems to be no way to get these expressions from both the methods to match. I know the arbitrary constant could vary in different methods but even the strcture of both these expressions are not ...
1
vote
2answers
565 views

Why does using u-substitution give a different answer? [duplicate]

I am working on a problem that required finding $\int (-7-y)^2dy$. Instead of foiling it out, I just used u-substitution to find the antiderivative. ($u=-7-y; u'=-1$). I got an antiderivative of $\...
7
votes
2answers
197 views

Evaluating $\int \sqrt{\frac{5-x}{x-2}}\,dx$ with two different methods and getting two different results [duplicate]

I tried Evaluating $\int \sqrt{\dfrac{5-x}{x-2}}dx$ using two different methods and got two different results. Getting two different answers when tried using two different methods:- M-$1$: $$\int \...
0
votes
2answers
69 views

Intergral of x/(2x-2) - Two different answers, which one is correct? [duplicate]

I want to find the integral $$\int\frac{x}{2x-2}dx$$ This is just a simple question from my textbook. But there seems to be two ways of solving it. If I simplify it to: $$\int1+\frac{2}{2x-2}dx$$ I ...
0
votes
2answers
241 views

Different answer for integral for two different methods [duplicate]

I am trying to integrate $\frac{1-x}{(x+1)^2}$, but I get to answers for two different methods: First, $\frac{1-x}{(x+1)^2} = \frac{1-x+1-1}{(x+1)^2} = \frac{2}{(x+1)^2} - \frac{x+1}{(x+1)^2} = \frac{...
0
votes
3answers
59 views

Integral of $\tan^3(x)$ [duplicate]

I'm going through Stewart's Calculus, and in section 7.3, example 7 asks for $\int \tan^3(x)dx$. In the book, the solution is $\frac{\tan^2(x)}{2} - \ln|\sec x| + C$. This was obtained in the ...
0
votes
3answers
70 views

Should I add $C$ (constant of integration) before or after calculation (or does it matter)? [duplicate]

For example, determine $\int \left(\frac{1}{2x+1}\right)dx$. Given that $f(x)$ = $\ln(2x+1)$ and $f'(x)$ = $\ln\left(\frac{2}{2x+1}\right)$. Would this be $\frac{1}{2} \int\left (\frac{2}{2x+1}\...
1
vote
1answer
59 views

Possible counter-proof of the constant rule of integration [duplicate]

$$∫(2x+10)^{-1}dx$$ $$=\frac{1}{2}∫(x+5)^{-1}dx$$ $$=\frac{1}{2}\ln|x+5|+C$$ but here is the same integral, yet different answer $$∫(2x+10)^{-1}dx$$ Let $u = 2x+10\implies \frac{du}{dx} = 2 \implies ...
0
votes
1answer
54 views

Does $-2 \int (\cos^2\,x-\sin^2\,x)\,\sin\,2x\,dx$ have multiple answers? [duplicate]

After working $-2 \displaystyle\int (\cos^2\,x-\sin^2\,x)\,\sin\,2x\,dx$, I got $\frac{-\sin^2(2x)}{2}+c$. The answer I was given was $\frac{\cos^2(2x)}{2}+c$. Are these both legitimate solutions, ...
52
votes
6answers
8k views

How can this function have two different antiderivatives?

I'm currently operating with the following integral: $$\int\frac{u'(t)}{(1-u(t))^2} dt$$ But I notice that $$\frac{d}{dt} \frac{u(t)}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$ and $$\frac{d}{dt} \frac{...
9
votes
3answers
4k views

Two different solutions to integral

Given the very simple integral \begin{equation} \int -\frac{1}{2x} dx \end{equation} The obvious solution is \begin{equation} \int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} ...
2
votes
3answers
454 views

Which of these answers is the correct indefinite integral? (Using trig-substitution or $u$-substitution give different answers)

Answers obtained from two online integral calculators: $$\begin{align}\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= -\sqrt{\dfrac{x + 1}{1 - x}} + \sqrt{\dfrac{x + 1}{1 - x}}x - 2\...

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