Questions tagged [indefinite-integrals]

Question about finding the primitives of a given function, whether or not elementary.

The indefinite integral is defined as a set of all functions $F$ such that $F' = f$. Each member of the set is called an antiderivative. For example, $$\int f(x) dx = \lbrace F(x): F'(x) = f(x) \rbrace$$ also commonly denoted as $$F(x) + C.$$

If $F'(z) = f(z)$ then we denote

$$\int f(z) \; dz = F(z)$$

and call $F(z)$ a primitive of $f(z)$, also called an antiderivative. This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.

Since the derivative of a constant is zero, any constant may be added to an antiderivative and will still correspond to the same integral. Another way of stating this is that the antiderivative is a nonunique inverse of the derivative. For this reason, indefinite integrals are often written in the form $$\int f(z)\;dz=F(z)+C$$

where $C$ is an arbitrary constant known as the constant of integration.

It may happen that there is no elementary function$^1$ such that $$\int f(z) \; dz = F(z)$$ In such case, we define a new function which is not elementary but still satisfies our definition. For example, there is no elementary function $F$ such that $F'(z) = \displaystyle \frac{e^z}{z}$. However, if we define

$$\int \frac{e^z}{z} dz = C + \log z + \int_0^z \frac{e^t-1}{t} dt$$

we can readily check that $F' = f$.

$^1$: A function built up of a finite combination of constant functions, field operations (addition, multiplication, division, and root extractions - the elementary operations) and algebraic, exponential, and logarithmic functions and their inverses under repeated compositions. See also.

Source: Wolfram Mathworld

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Is there any solution to this indefinite integral

How to solve the following indefinite integral? $$ \int \frac{\lambda \alpha t^\alpha}{(1+\lambda t^{\alpha})^2}dt $$ where $\lambda,\alpha$ are real numbers. I've tried to integration by part, however I came across $\int \frac{1}{1+\lambda…
sss1031
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Help in finding the integral function.

Can somebody provide a hint in finding the following integral? $$\displaystyle \int \dfrac{1}{(x^3+1)^3} \text{ d}x$$ I thought of using partial fractions but that isn't making any sense.
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What to do next with this indefinite integral?

I'd like to evaluate the following indefinite integral: $$\displaystyle\int \frac{dx}{(3+x^2)(\sqrt{1+x})}$$ I started by letting $y^2=1+x$ and, after simplifying, got here: $$ 2\int\frac{dy}{y^4-2y^2+4}$$ I'm not sure what to do next. Does anyone…
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Indefinite integral: $\int \frac{x\mathrm{ln}x}{(1+x^2)^2}\mathrm{d}x$

I am stuck with the following integral: $$\int \frac{x\mathrm{ln}x}{(1+x^2)^2}\mathrm{d}x$$ I'm sure integration by parts has to be involved here, but I cannot find the proper function to use the formula. Any ideas would be helpful.
user209217
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What is indefinite $ \int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - x^{4}}}{\rm d}x . $

$$ \int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - x^{4}}}{\rm d}x . $$ Hello. Today, in exam, we had to evaluate this integral. Noone was able to do it. Appreciate any help.
Pompo
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Finding $ \int \sin^2(2x)/[1+\cos(2x)]dx$.

I am surprisingly having a bit of difficulty with an indefinite integral which is interesting since the integral I solved before is $$ \int \frac{1+\cos2x}{\sin^2(2x)} dx$$ The integral I am currently working on is $$ \int…
user156926
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integral substitution that use the derivative

I tried to solve the equation $$\frac{dy}{dx}=\frac{k}{\sin{x}\sqrt{\sin^2{x}-k^2}}$$ and was suggested the substitution $$ \tan{u}=\frac{dy}{dx}\sin{x} $$ after some algebra (I'll add it if necessary) I get $$ \frac{dy}{du} =…
Michael
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Find $\int{\frac{1}{\left(1+\ln x\right)^2}\;dx}$

How would you integrate a function almost entirely in logarithmic form, such as:$$\int{\frac{1}{\left(1+\ln x\right)^2}\;dx}$$ I have tried various substitutions and considered integrating by parts, but I can't seem to get a quick and elegant…
Hernandez
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Apparently inoffensive indefinite integral. (grad student here)

I've been dealing with these for a while, and tried different things with no success as of yet: $$ \int \frac {dx}{(x²+c²)\sqrt{(x-a)(x-b)}} $$ $a$, $b$ and $c$ are real positive numbers. Trying solving it in Mathematica, if $c=0$, the answer is…
magtovi
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how to calculate this indefinite integral?

I'm studying for an exam and I found this excercise in my textbook, but I don't know how to evaluate this integral (I've never seen one with exponentials like that), any hint? $\displaystyle\int 2^{x^3-x^2}(6x^2-4x)$ I noticed that part of the…
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Express $\int\exp\left(\frac{a}{x}+bx\right)x^{\eta}\mathrm{d}x$ in terms of special functions?

Can the following integral: $$\int\exp\left(\frac{a}{x}+bx\right)x^{\eta}\mathrm{d}x$$ be expressed in terms of special functions, like the Gaussian hyper geometric function $_2F_1(a,b;c;z)$, or the incomplete Gamma function? Ideally, use only…
a06e
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Is brute force the only answer to $\int u^2(1-u^2)^4du$?

Feedback: After some initial interest, it seems this question has been down voted. Jyrki Lahtonen may be right that spending time on unsolvable problems is pointless, but, to a student, at what point does it become clear that it's pointless? I…
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Evaluation of $\displaystyle \int \frac{x^2}{(x^4-1)\sqrt{x^4+1}}dx$

Evaluation of $\displaystyle \int \frac{x^2}{(x^4-1)\sqrt{x^4+1}}dx$ What I try : Using Substution, $\displaystyle x=\frac{1-t}{1+t}$ and $\displaystyle dx =\frac{2}{(1+t)^2}dt$ also using $\displaystyle x^4-1=\frac{2(1+6t^2+t^4}{(1+t)^4}dt$ And…
jacky
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integral $I=\int\dfrac{dx}{x\sqrt{2+x-x^2}}$

I am trying to calculate the following integral $$I=\int\dfrac{dx}{x\sqrt{2+x-x^2}}$$ The answer in my textbook is $$-\dfrac{1}{\sqrt2}\ln\bigl|\dfrac{\sqrt{2+x-x^2}+\sqrt2}{x}+\dfrac{1}{2\sqrt2}\bigr|$$ Note that…
SAQ
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Calculate $I=\int \sqrt{x^2-a^2} dx$ with change of variable

Let's try to calculate the integral $$I=\int \sqrt{x^2-a^2} dx, x>a>0$$ I think a suitable substitution would be $x=\dfrac{a}{\sin t}$. Then we would have $$dx=d\dfrac{a}{\sin t}=ad\frac{1}{\sin t}=ad\sin^{-1}t=-1a\sin^{-2}t\cos tdt=-\dfrac{a\cos…
kormoran
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