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

5544 questions
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Integration of elementary function by shortest possible method.

$$\int {\frac{1}{x^\frac13+x^\frac14}+\frac{log(1+x^\frac16)}{x^\frac13+x^\frac12}}dx$$ I have to sovle this function integral. Please help me out I have tried a lot but its approach is very long. I have also seen its solution bit still its quite…
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Tricky Integral Problem with tan and sec function

Can someone help me evaluate:$$\int \frac{(\sec x)^{2}}{(1+\tan x)^{2}}dx$$ Is it possible for a hint so that I can proceed? I tried changing sec into $ 1 +\tan x $ but did not reach far.
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Find the following indefinite integral including a root and fraction

I'm trying to solve the following indefinite integral. I know I need to do substitution, but I simply don't know how to do it. $$\int \:\frac{x^2}{\sqrt{\left(x^2+1\right)^3}}$$
DitJid
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$\int\frac {dx}{\sqrt{2ax−x²}}=a^n \sin^{-1}(\frac {x}{a}-1)$

The question is telling us to find n The options are: a) 0 b) -1 c) 1 d) none of these I have tried to solve this by using some general formula of integration yet Iam unable to find the answer
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Integral calculus .

This is a wonderful integral .already been able to find some steps to solve it but always incomplete . The integral is stated below. $$ \int_0^1 \frac{\mbox{arcsinh}(x)}{1+x^2} dx$$ Need a nice soln , i have a wonderful feeling it will be…
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How to solve for dx in the numerator by itself

$$\int \frac{dx}{1+x^2} $$ I'm having a hard time knowing what to do when it's just dx on top
A.Bree
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Evaluating $\int\frac{\sqrt{1-x}}{x}\,dx$

$$\int\frac{\sqrt{1-x}}{x}\,dx$$ $$\int \:uv'=uv-\int \:u'v$$ $$u=\frac{\sqrt{1-x}}{x},\:\:u'=\frac{x-2}{2\sqrt{1-x}x^2},\:\:v'=1,\:\:v=x$$
luke
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Evaluate the following indefinite integral. $\int\frac{1}{x^4+1}\, dx$

I would like to evaluate the following indefinite integral: $\int\frac{1}{x^4+1}\, dx$
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cleopatra's problems

Evaluate $\int{\frac{9x+6}{(3x-2)^4(x+2)^4}dx}$
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Find $\int\sin^3(2x+1)dx$.

Find $\int\sin^3(2x+1)dx$. Having three different results which one is right right? 1.$y= -\frac{\cos (2x+1)}{2} +\frac{1}{24\cos(6x^3)} -\frac{1}{4\cos(2x+1)}+c$ $y=-\cos (2x+1) + \cos^3(2x+1) +c.$ $y= -\frac{\cos(6x+3)}{24} …
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What is a triple integral with the answer of 39?

What is a triple integral with the answer of 39
Angela
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