If a derivative tells you how fast something changes, an antiderivative recovers the original quantity from its rate of change. Given f, we look for F with F′(x)=f(x). The collection of all such F is the indefinite integral, written ∫f(x)dx. It is a family of functions , any two members differ only by an additive constant , because differentiation kills constants.
Definitions
An antiderivative of f on an interval I is a function F such that F′(x)=f(x) for every x∈I. The indefinite integral is
∫f(x)dx=F(x)+C,
where C∈R is the constant of integration. The symbol ∫ comes from a stretched "S" (for sum), f(x) is the integrand, and dx indicates the variable of integration.
Why the "+C"?
Suppose F′(x)=G′(x)=f(x) on an interval. Then (F−G)′(x)=0, so F−G is constant on that interval. Hence any two antiderivatives differ by a constant. We capture all of them by writing F(x)+C.
The constant is more than a formality. In applied problems , e.g. recovering position from velocity , the constant encodes the initial condition. Without it, the answer is ambiguous. Always write the +C in indefinite integrals.
Linearity
Integration distributes over sums and pulls out constants:
∫[af(x)+bg(x)]dx=a∫f(x)dx+b∫g(x)dx.
This is the workhorse for splitting a complicated integrand into easy pieces. The identity is immediate from the corresponding property of derivatives, applied in reverse.
Note that integration does not play nicely with products or quotients. There is no simple "integral of a product" rule , you must use integration by parts. There is no quotient rule for integrals , you may need partial fractions, substitution, or other tools.
The standard integral table
Every entry below is a derivative formula read backwards. Verify each one by differentiating the right side.
∫f(x)dx
Result
∫xndx, n=−1
n+1xn+1+C
∫xdx
$\ln
∫exdx
ex+C
∫axdx
lnaax+C
∫sinxdx
−cosx+C
∫cosxdx
sinx+C
∫sec2xdx
tanx+C
∫csc2xdx
−cotx+C
∫secxtanxdx
secx+C
∫cscxcotxdx
−cscx+C
∫tanxdx
$\ln
∫cotxdx
$\ln
∫secxdx
$\ln
∫cscxdx
$\ln
∫1+x2dx
tan−1x+C
∫1−x2dx
sin−1x+C
∫xx2−1dx
$\sec^{-1}
A few remarks. The integral of sinx is −cosx (mind the minus sign , a common error). ∫tanxdx may equivalently be written as −ln∣cosx∣+C, which is the form some prefer; both are correct. The formula for ∫secxdx looks mysterious , it is derived by multiplying numerator and denominator by secx+tanx and recognising a u-substitution.
Worked examples
Example 1. Compute ∫(3x2−4sinx+ex)dx.
By linearity: ∫3x2dx=x3, ∫−4sinxdx=4cosx, ∫exdx=ex. So the integral is x3+4cosx+ex+C.
Example 2. Compute ∫xx2+1dx.
Split: xx2+1=x+x1. Integrate term by term: 2x2+ln∣x∣+C.
Example 3. Compute ∫21−cos2xdx.
Recognise 21−cos2x=sin2x. But it is easier to integrate directly: ∫21dx−∫2cos2xdx=2x−4sin2x+C.
Example 4. Compute ∫tan2xdx.
Use tan2x=sec2x−1. Then ∫tan2xdx=tanx−x+C.
Example 5. Find F(x) with F′(x)=3x2+2 and F(1)=5.
Antiderivative: F(x)=x3+2x+C. Apply the condition: 1+2+C=5, so C=2. Therefore F(x)=x3+2x+2.
Example 6. Compute ∫(ex+5)2dx.
Expand: e2x+10ex+25. Integrate: 2e2x+10ex+25x+C.
Try it yourself
∫(5x4−3x2+7)dx
∫x2x3−2x+1dx
∫cos2xdx (use cos2x=21+cos2x)
∫sin2xdx
∫(secx+tanx)secxdx
∫cos2x1+sinxdx
∫3xdx
∫(ex−1)2exdx (hint: anticipate substitution but try expanding first)
∫(x+1)2dx
Find f if f′(x)=cosx+sec2x and f(0)=1.
∫cot2xdx
∫sin2x1−cosxdx (use 1−cosx=2sin2(x/2))
∫x−3/2dx
Verify ∫secxdx=ln∣secx+tanx∣+C by differentiating the right side.
Pitfalls and tricks
Forgetting +C: half a mark in board exams, every time.
Sign on ∫sinxdx: the answer is −cosx, not cosx.
∫xdx is ln∣x∣, not lnx , the absolute value matters because the function is defined for x<0 too.
The power rule fails at n=−1 , that case is the log.
Simplify first: many integrands look hopeless until you split a fraction, expand a square, or use a trigonometric identity. Always try algebraic simplification before reaching for a heavy technique.