# Antiderivative and Indefinite Integration

In calculus, the **antiderivative** of a function \(f(x)\) is a function \(F(x)\) such that \( \frac{dF(x)}{dx} = f(x) + C.\) That is, the derivative of \(F(x)\) is \(f(x).\) This is also known as the **indefinite integral**. The constant \(C\) is called the constant of integration. This identity is the first part of the fundamental theorem of calculus.

Why must we have the constant of integration? Think about it. If you have differentiated something and get the result \( 2x \), how many things could you have differentiated to get \( 2x \)? An infinite amount: \( x^2, x^2 + 1, x^2 + 2, x^2 + 2016, \) etc.

There is no pre-defined method for computing indefinite integrals, so several different integration techniques have been developed.

## Find the antiderivative of \(56x^{55}.\)

We notice that the derivative of \(x^{56}\) is \(56x^{56-1}=56x^{55}\), the antiderivative of \(56x^{55}\) would be \(x^{56}.\)

We can also evaluate the polynomial using normal integration: \[\int 56x^{55}\, dx= 56\int x ^ {55} dx= \frac{56}{55+1} x^{55+1} +C= x^{56}+C, \] where \(C\) is the constant of integration. \(_\square\)

## If \(f(x)=\cos x\), what is the antiderivative of \(f(x)?\)

From the limit definition of the derivative we know that \(\frac{d}{dx}\sin x=\cos x\). Since \(\cos x\) is the derivative of \(\sin x\), from the definition of antiderivative, the antiderivative of \(\cos x\) must be \(\sin x\) plus the constant of integration \(C\). Thus, the integral of \(\cos x\) must be \(\sin x\). \(_\square\)

## Find the anti-derivative of \(\dfrac 3{\sqrt x}+\dfrac{1}{2}\sin x.\)

We see that the derivative of \(2\sqrt x\) is \(\dfrac{1}{\sqrt x}\). Therefore, \[\displaystyle\int \dfrac 3{\sqrt x}\, dx= 3\times(2\sqrt x)+C=6\sqrt x + C,\] where \(C\) is the constant of integration.

Also, we find that the derivative of \(-\dfrac 12\cos x\) is \(\dfrac{1}{2}\sin x\). Thus, \[\int\left( \dfrac 3{\sqrt x}+\dfrac{1}{2}\sin x\right) \, dx = 6\sqrt x -\dfrac 12\cos x + C. \ _\square\]

## If \( f(x) = xe^{x^2} \), what is \(\displaystyle \int f(x)\,dx? \)

We use u-substitution here.

Substituting \(x^2\) for \(u\), we get \(du = 2x\, dx\) and \(x = \frac{du}{2dx}.\) From the above we get \[ \frac{1}{2} \int e^u\, du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C,\] where \(C\) is the constant of integration. \(_\square\)

Evaluate \( \displaystyle \int \dfrac{1+ \sin x }{\cos^{2} x} \, dx. \)

Breaking up the integrand gives \[\begin{align} \int \dfrac{1+ \sin x }{\cos^{2} x} \, dx &= \int \left( \dfrac{1}{\cos^{2} x } + \dfrac{\sin x }{\cos^{2} x} \right) \, dx \\ &= \int \left( \sec^{2} x + \sec x \tan x \right) \, dx. \end{align} \] Integrating term by term, we have familiar results as follows: \[ \int \left( \sec^{2} x + \sec x \tan x \right) \, dx= \tan x + \sec x + C, \] where \(C\) is the constant of integration. \(_\square\)

Evaluate \( \displaystyle \int x \cos x \, dx. \)

For this indefinite integral, we need to use integration by parts. According to the LIATE, set \( u = x \Rightarrow \dfrac{du}{dx} =1\) and \(v = \sin x \Rightarrow \dfrac{dv}{dx}=\cos x. \) Then

\[\begin{align} \int x \cos x \, dx &= x \sin x - \int \sin x \, dx \\ &= x \sin x + \cos x + C, \end{align}\]

where \(C\) is the constant of integration. \(_\square\)

Evaluate \( \displaystyle \int \tan^{3} x \sec^{2} x \, dx. \)

Here, we should use the u-substitution \( u = \tan x \). Then \( \dfrac{du}{dx} = \sec^{2}x \), which goes away. Then the given expression is equal to

\[\begin{align} \int \tan^{3} x \sec^{2} x \, dx &=\int u^{3} \, du\\ &=\dfrac{u^4}{4} + C \\ &= \dfrac{\tan^4 x}{4} + C, \end{align} \] where \(C\) is the constant of integration. \(_\square\)

Evaluate \( \displaystyle \int \dfrac{dx}{x^{2} + 3x + 2 } \).

For this problem, we should do a partial fraction decomposition.

The integrand is \( \dfrac{1}{(x+1)(x+2)} = \dfrac{A}{x+1} + \dfrac{B}{x+2 }, \) which gives \( A(x+2) + B(x+1) = 1. \)

Let \( x = -2 \), then \( B = -1. \)

Let \(x = -1 \), then \( A = 1.\)Then we can rewrite the integrand as \( \dfrac{1}{x + 2 } - \dfrac{1}{x+1} \).

Using u-substitution, we can get the answer \( \ln \left| x+ 2 \right| - \ln \left| x+1 \right| + C \), which can alternatively be written as \[ \ln \left| \dfrac{x+2}{x+1}\right| +C,\] where \(C\) is the constant of integration. \(_\square\)

**Cite as:**Antiderivative and Indefinite Integration.

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