Indefinite Definition and 299 Threads

Evaluation of $$\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx$$$\bf{My\;Try::}$ Let $\displaystyle f(x) = \frac{ax+b}{(x^3+3x+1)^2}.$ Now Diff. both side w. r to $x\;,$ We Get$\displaystyle \Rightarrow f'(x) = \left\{\frac{(x^3+3x+1)^2\cdot a-2\cdot (x^3+3x+1)\cdot (3x^2+3)\cdot...

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I actually came across this question on social media. What is: $$\int sin (x) \, dx - \int sin (x) \, dx$$ And I think the answer depends on how we interpret: $$\int sin (x) \, dx$$ If we think of it as a single antiderivative, the answer would be zero. If we think of it as being...

Hello, everyone I've trying to solve this integral but it seems like the methods I know are not enogh to solve it. So I'd be glad if you could give me some trick to get into the answer. Here it is: Thanks in advance!

Several authors state the formula for finding the arc length of a curve defined by ##y = f(x)## from ##x=a## to ##x=b## as: $$\int ds = \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx$$ Isn't this notation technically wrong, since the RHS is a definite integral, and the LHS is an indefinite integral...

Hello, I was just wondering, we have what could be called the indefinite derivative in the form of d/dx x^2=2x & evaluating at a particular x to get the definite derivative at that x. But with derivation, we can algebraically manipulate the limit definition of a derivative to actually evaluate...

Argument A ##∫ 0 dx = 0x + C = C## Argument B ##∫ 0 dx = ∫ (0)(1) dx = 0 ∫ 1 dx = 0(x+C) = 0## Discuss.

\displaystyle \int e^{x^4}\left(x+x^3+2x^5\right)\cdot e^{x^2}dx =

Calculation of $\displaystyle \int \frac{\sqrt[4]{x^{10}+x^8+1}}{x^6}\cdot \left(3x^{10}+2x^{8}-2\right)dx$ $\bf{My\; Try::} $ Given $\displaystyle \int \frac{\sqrt[4]{x^{10}+x^8+1}}{x^6}\cdot \left(3x^{10}+2x^{8}-2\right)dx = \int\frac{\sqrt{x^6+x^4+x^{-4}}}{x^5}\cdot...

The big clue here is the square root in the denominator, because $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{x} \right) = \frac{1}{2\,\sqrt{x}} \end{align*}$. So this suggests that you probably need to find a square root function to substitute. Rewrite your...

Calculation of \(\displaystyle \int\frac{\sqrt{\cos 2x}}{\sin x}dx\) **My Try \(\displaystyle :: I = \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \int\frac{\cos 2x}{\sin x\cdot \sqrt{\cos 2x}}\cdot \frac{\sin x}{\sin x}dx = \int\frac{\sqrt{2\cos^2 x-1}}{\left(1-\cos^2 x\right)\cdot \sqrt{2\cos^2...

What is the indefinite integral of cosec(\theta)?

Is possible to compute indefinite integrals of functions wrt its variables, but is possible to compute indefinite integrals of scalar fields and vector fields wrt line, area, surface and volume?

Hi, I tried to solve this integral \int\sqrt{1-\frac{1}{x^3}}dx but i can't solve it... can someone help me?

Is this possible.. Say, a∫b f(x)dx = G(x)|x=b - G(x)|x=a = S, where S, a and b are known. Can we find G(x) ?

Evaluation of $\displaystyle \int\sqrt\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx$ I have Tried The Given Integral Using $\displaystyle \tan x = \frac{2\tan \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ and $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ and $\displaystyle \sin x...

$\displaystyle \int \frac{\ln\left(x^2+2\right)}{(x+2)^2}dx$ $\bf{My\; Try::}$ Given $\displaystyle \int \ln \left(x^2+2\right)\cdot \frac{1}{(x+2)^2}dx$ Using Integration by parts, we get $\displaystyle = -\ln\left(x^2+2\right)\cdot \frac{1}{(x+2)} + 2\int \frac{x}{\left(x^2+2\right)\cdot...

Homework Statement Evaluate the indefinite integral of x*cos(3x)^2 Homework Equations Integration by parts: \int(udv)= uv - \int(vdu) The Attempt at a Solution Im having trouble finding the antiderivative of cos(3x)^2 (which I designated as dv when doing integration by parts). I...

Homework Statement Integrate the following indefinite integrals A:\int e^x (x^2+1) dx B:\int e^x cos(3x+2) dxHomework Equations \int u dv = uv - \int v du The Attempt at a Solution Part A: I have done the following but when I use an integration calculator online its not what I have (although...

Indefinite integrals with different solutions? Homework Statement \int \csc ^{2}2x\cot 2x\: dx Solve without substitution using pattern recognition Homework Equations As above The Attempt at a Solution To try a function that, when differentiated, is of the same form as the...

How does \int \frac{2x + 2}{x^2 + 2x + 5} \, dx turn into \ln(x^2 + 2x + 5)? How are they getting rid of the numerator are they just dividing by the reciprocal of 2x + 2?

in understand why we write the dx in riemann integral , but in the indefinite integral why do we use that ? what is the relation between the area under a curve , and the antiderivative of that of that curve ??

Okay so I'm working on this problem \int \frac{x^2}{\sqrt{4 - x^2}} \, dx I do a substitution and set x={\sqrt{4}}sinu I get to this step fine \int 4sin(u)^2 I know that u = arcsin(x/2) so I don't see why I can't just substitute in u into sin(u)? I tried this and I got \int 4 *...

I understand why a definite integral of the form ^{b}_{a}∫ƒ(x)dx has the differential dx in it. What I don't understand, and what my teacher hasn't explained is why an indefinite integral (i.e. an antiderivative) requires the differential. Why does ∫ƒ(x)dx require that dx to mean...

Here is the question: I have posted a link there to this thread so the OP can view my work.

Homework Statement Solve: \int sin(16x) \sqrt[a]{cos(16x)}\,dx Answer should be linear in the constant "a" The Attempt at a Solution \int sin(16x) \sqrt[a]{cos(16x)}\,dx Set: u=cos(16x), du=-16sin(16x) du ~~\Rightarrow~~ {-1/16}\int \sqrt[a]{u}\,du =...

Here is the question: I have posted a link there to this thread so the OP can view my work.

I this old thread it mentions that the indefinite integral of f'(x)/f(x) is log(|f(x)|)+C which means that there is some ambiguity about the sign of f(x). There does however, seem to be no ambiguity about the value of C as it always appears to be zero, but I have never seen this mentioned...

Here are the questions: I have posted a link there to this thread so the OP can see my work.

I am trying to find the following indefinite integral: Homework Statement ∫\sqrt{}x/(x-1)dx Homework Equations None The Attempt at a Solution I tried to use substitution but got nowhere. I set u=\sqrt{}x so du=1/(2\sqrt{}x)dx. However from here on on I got stuck. I also tried...

Here is the question: I have posted a link there to this thread so the OP can see my work.

Here is the question: I have posted a link there to this topic so the OP can see my work.

[1] $\displaystyle \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}dx$ [2] $\displaystyle \int \frac{3\cot 3x - \cot x}{\tan x-3 \tan 3x}dx$ Thanks pranav I have edited it.

$\displaystyle \int\frac{1}{(a+b\sin x)^2}dx$, where $a>b$ My Trial :: Using Integration by parts:: $\displaystyle \int\frac{1}{(a+b\sin x)^2}dx = \frac{1}{b}\int -\csc (x)\cdot \frac{-b\sin x}{(a+b\sin x)^2}dx$ $\displaystyle -\frac{1}{b}\cdot \csc (x)\cdot \frac{-1}{(a+b\sin x)}+\int...

I'm trying to integrate \int e^{4\ln{x}}x^2 dx I can't see using u-substition, x^2 isn't the derivative of e^{4\ln{x}} nor vice-versa. I tried integrating by parts and that didn't work. I used u=e^{4\ln{x}} and dv=x^2 dx I know I can't rewrite e^{4\ln{x}} as e^4e^\ln{x}

$(a)\;\;:: \displaystyle \int\frac{1}{\left(x+\sqrt{x\cdot (x+1)}\right)^2}dx$ $(b)\;\;::\displaystyle \int\frac{1}{(x^4-1)^2}dx$ My Trial :: (a) $\displaystyle \int\frac{1}{(x+\sqrt{x\cdot (x+1)})^2}dx$ $\displaystyle \int\frac{1}{x\left(\sqrt{x}+\sqrt{x+1}\right)^2}dx =...

how would start solving this $\displaystyle\int\frac{dy}{1+e^y}$

Problem: Rewrite the indefinite integral ## \iint\limits_R\, (x+y) dx \ dy ## in terms of elliptic coordinates ##(u,v)##, where ## x=acosh(u)cos(v) ## and ## y=asinh(u)sin(v) ##. Attempt at a Solution: So would it be something like, ## \iint\limits_R\, (x+y) dx \ dy =...

Homework Statement Sorry for the poor use of Latex, I have tried to get it to work but it seems to never come out as I would like. Using a trigonometric or hyperbolic substitution, evaluate the following indfenite integral, ∫\frac{1}{\sqrt{(x^2-1)^5}} dx Homework Equations I...

how would i go about solving this $\displaystyle\int\frac{\arctan x}{1+x^2}dx$? i tried substitution but i didn't work.

Here is the question: I have posted a link there to this topic so the OP can see my work.

Here is the question: I have posted a link there to this topic so the OP can see my work.

Here is the question: I have posted a link there to this topic so the OP can see my work.

This is the integral I am trying to evaluate. I would very much appreciate any help. \[\int \frac{2x^3-1}{x+x^4}dx\] MY APPROACH: \[\int \frac{2x^3-1}{x+x^4}dx\] \[\int \frac{1}{2}.(\frac{4x^3+1}{x^4+x}-\frac{3}{x^4+x})dx\] \[\frac{1}{2}\log{x+x^4}-\frac{1}{2}\int \frac{3}{x^4+x})dx\] now we...

Homework Statement Ok the problem is: ∫-1/(4x-x^2) dx The answer in the back of the book is: (1/4)ln(abs((x-4)/x))) + C Homework Equations I think this would be used somehow: ∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + C The Attempt at a Solution ∫-1/(4x-x^2) dx...

Here is the question: I have posted a link there to this topic so the OP can see my work.

Homework Statement ∫1/(t*ln(t)) dt ∫1/(√(t)*[1-2*√(t)]) dt Homework Equations The Attempt at a Solution I used u-substitution for both. For the first equation, my u= ln t, and my final answer was ln|u| + C, or ln(ln(|t|) + C. For the second equation, my u= 1-2*√(t) and...

Homework Statement \int \left(\frac{x}{x\cos x-\sin x} \right)^2 dxHomework Equations The Attempt at a Solution Factoring out ##\cos x## from the denominator, the integral transforms to \int \sec^2x \left(\frac{x}{x-\tan x}\right)^2dx Substituting ##\tan x=t##, ##\sec^2 xdx=dt## \int...

Homework Statement \int \sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}} \frac{\sec x}{\sqrt{1+2\sec x}}dxHomework Equations The Attempt at a Solution The integral can be simplified to: \int \sqrt{\frac{1-\cos x}{1+\cos x}} \frac{1}{\sqrt{\cos x} \sqrt{\cos x+2}}dx Using ##\cos...

Homework Statement \int1/(1+\sqrt{2x})\,dx Homework Equations u=1+\sqrt{2x} \Rightarrow \sqrt{2x}=u-1 du=1/\sqrt{2x}dx \Rightarrow \sqrt{2x}du=dxThe Attempt at a Solution \int1/(1+\sqrt{2x})\,dx = \int\sqrt{2x}/(1+\sqrt{2x})\,du = \int(u-1)/u\,du = \int\,du-\int1/u\,du = u-ln|u|+C =...