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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How to find the antiderivative of the $\sqrt{4-x^2}$?

How to find the $\int \sqrt{4-x^2}$ I know that you can use substitution for x and sub in $2\sin\theta$, but i don't understand how and why you know to do that? That is not something I would think of doing
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$\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$

I got this question $\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$ and I solved it as thus: $$\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$$ $$=\int \frac{6\sin x…
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What's $ \int \frac{1}{2+\cos 2x}dx$ on earth?

Let's consider a problem, which is to find the indefinite integral $$I(x):=\displaystyle \int \frac{1}{2+\cos 2x}dx.$$ Since the integrand $f(x):=\dfrac{1}{2+\cos 2x}$ is continuous over $(-\infty,+\infty)$, hence $I(x)$ exists necessarily,…
mengdie1982
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Does $\int \frac{e^x(x-1)}{1+xe^x}dx$ have any closed form?

It's simple that $$\int \frac{e^x(x+1)}{1+xe^x}dx=\ln(1+xe^x)+C,$$ But what if $$\int \frac{e^x(x-1)}{1+xe^x}dx?$$
mengdie1982
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Why can't I make the substitution $ u = \sin (ax + b) $ to evaluate $ \int \sin (ax + b) \cos (ax + b) dx$?

Evaluate $ \int \sin (ax + b) \cos (ax + b) dx$? To do this, I started of by substituting $ u = \sin (ax + b) $. That made $ du = cos (ax + b) \cdot a $ and wrote the integral as $ \frac 1a \int u \ du $ to get the final answer as: $$ \frac 1{2a}…
WorldGov
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Indefinite integral $\int \ln(1-x)dx$

When asked to find the indefinite integral $$\int \ln(1-x)dx$$ my first instinct is to use a u-substitution and let $u=1-x$ and therefore $-du=dx$ And after the substitution we have the indefinite integral with u as the variable of integration…
Isosceles
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Evaluate $ \int {(\tan x)}^{\frac{1}{3}} dx $

Evaluate the following indefinite integral. $$ \int {(\tan x)}^{\frac{1}{3}} dx $$ How could i do this integral ? I tried multiplying the numerator by ${(\sec x)}^2$ and denominator by $1+{(\tan x)}^2$ but getting stuck after putting $\tan x=t$.
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Integral of greatest integer function.

$[x]$ denotes the greatest integer $\leq x$. Let $f(x)=[x]$ What is the indefinite integral (without limits) of $f(x)$ ? And if $f(x) = e^{-7[x]}$ what would the indefinite integral be?
Gin99
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Integrate $ \ \int \frac{dv}{4v+\ln v} $

Integrate :$$ \ \int \frac{dv}{4v+\ln v}$$ I have tried in the following way- $ \ \\ Let \ \ \ln v=z , \\ or, dv=vdz , \\ or, dv=e^{z} dz $ $$ $$ Hence, $$ \ \int \frac{dv}{4v+\ln v} = \int \frac{e^{z}}{4e^{z}+z} dz = \frac{1}{4} \int…
MAS
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How to do this without using trigonometric substitution?

$$\int \frac {\rho^2}{(\rho^2+ h^2)^\frac 32} d\rho$$
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Evaluate the integral

I need some help in evaluating this integral: $$\int {\frac{{\cos x}}{{{{\sin }^3}x + \sin x + 4}}dx} $$ I've tried using the substitution $u=\sin{x}$ but I ended up with a cubic polynomial in the denominator.
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Just another problem on Integration:-

What I tried was I took $$\tan^{-1}(x) =t, $$ but I got terms like $$\cos(\tan(t))$$ which I don't know what to do with. So please guide me into solving this..
satyatech
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Integral of x^2(x+6)^-6 without partial fractions

I have tried to solve this integral $$\int\frac{x^2}{(x+6)^6}\,\mathrm dx$$ by using partial fractions.
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How to solve the given integral

I didnt now how to solve the integral: $$\int\frac{\frac{\ln(a+bx)}{x}}{a+bx}dx$$ Please help me. Thanks for your help
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Primitivation and trigonometry: $\int \sin(\ln x)dx $

How to find the primitive of this guy by parts? I tried: $$\int \sin(\ln x)dx $$ $$=\int\cos(\ln x)*\tan(\ln x)dx$$ $$=\frac{1}{x*\cos(\ln x)}+\int\frac{1}{x*\cos^2(\ln x)}*\frac{1}{x}\sin(\ln x)dx$$ Is this the way to go?