Complete and check the integration
An answer you can test
The top curve shows the integrand \(\dfrac{x}{(x^2+3)^2}\). The teal curve below is a candidate antiderivative \(F\), which is an answer that you want to test.
You find it in two steps. First substitute \(u = x^2 + 3\), so that \(x\,dx = \tfrac12\,du\), and integrate in \(u\). Then replace \(u\) with \(x^2 + 3\), which gives \(F(x) = -\dfrac{1}{2(x^2+3)}\).
Each short orange dash below is as steep as the top curve is high at that x, and the teal curve runs along every dash. Drag the point. The orange bar on top is the height of the integrand, and the orange tangent below is the slope of F.
The two readouts stay equal. At \(x = 1\) both are \(\tfrac{1}{16}\).
At every x, the slope of an antiderivative equals the height of the integrand, so you test an answer by differentiating it.
Any answer
The top curve is any integrand \(f\), and the teal curve below is an antiderivative \(F\) plus a constant \(C\).
Drag the point. The slope of F equals the height of the integrand at the point, so F runs along the dashes.
Move the C slider. The teal curve moves up and down, but its slope at every x does not change. Every constant \(C\) therefore gives an antiderivative, and the answer is the whole family \(F(x) + C\).
The teal bracket is the rise of \(F\) from \(a\) to \(b\), the value of a definite integral. It is the same for every \(C\), because the constant cancels.
Notation
Integrate in \(u\). The result is an antiderivative in \(u\), plus a constant.
\[\int f(u)\,du = F(u) + C\]The problem is in \(x\), so replace \(u\) with \(g(x)\). The final answer must have no \(u\) in it.
\[\begin{aligned} &\int f(g(x))\,g'(x)\,dx \\ &\quad = \cy{F(g(x))} + C \end{aligned}\]Write one constant \(C\) at the end of the answer. Do not write a constant for each term.
For a definite integral, evaluate \(F(u)\) between the new limits. Or return to \(x\) and evaluate \(F(g(x))\) between the old limits. Both give the same number, the rise that the teal bracket shows.
To test an answer, differentiate it with the Chain Rule. The result must be the original integrand.
\[\begin{aligned} \dfrac{d}{dx}\,F(g(x)) &= F'(g(x))\,g'(x) \\ &= \cy{f(g(x))\,g'(x)} \end{aligned}\]
When it fails
Press the button to drop the factor \(\tfrac12\). The candidate becomes \(-\dfrac{1}{x^2+3}\).
Its slope at \(x = 1\) is \(\tfrac18\), but the integrand is only \(\tfrac{1}{16}\) high there. The red tangent is twice as steep as the orange one at every \(x\), and every dash turns red. A missing factor shows as a slope that does not match.
Example 1: back to x
Find \(\displaystyle\int 4x^3\sin(x^4)\,dx\) and test the answer.
The expression \(x^4\) sits inside the sine, and its differential is the rest of the integrand.
\[u = x^4, \qquad du = 4x^3\,dx\]Rewrite the integral and integrate in \(u\).
\[\int \sin u\,du = -\cos u + C\]Replace \(u\) with \(x^4\). The answer is now in \(x\), as the problem asked.
\[\cy{-\cos(x^4)} + C\]Differentiate the answer. The Chain Rule multiplies by \(4x^3\), the derivative of the inside function, and the result is the integrand.
\[\dfrac{d}{dx}\left[-\cos(x^4)\right] = \sin(x^4)\cdot 4x^3\]
Example 2: a scale factor at the end
Find \(\displaystyle\int x\cos(3x^2)\,dx\).
The expression \(3x^2\) sits inside the cosine. Its differential is \(6x\,dx\), so \(x\,dx\) is one sixth of \(du\).
\[u = 3x^2, \qquad x\,dx = \dfrac{1}{6}\,du\]Integrate in \(u\). The factor \(\tfrac16\) stays in front.
\[\dfrac{1}{6}\int \cos u\,du = \dfrac{1}{6}\sin u + C\]A tempting wrong move: Replace \(u\) and drop the factor \(\tfrac16\), because a constant can go into \(C\).
\[\cr{\sin(3x^2)} + C\]The factor \(\tfrac16\) multiplies the answer, so \(C\) cannot absorb it. The slope of \(\sin(3x^2)\) is \(6x\cos(3x^2)\), 6 times the integrand, and the red tangent is 6 times as steep as the orange one.
Keep the factor. The slope of this answer is the integrand.
\[\begin{aligned} &\cy{\dfrac{1}{6}\sin(3x^2)} + C \\ &\dfrac{d}{dx}\left[\dfrac{1}{6}\sin(3x^2)\right] = x\cos(3x^2) \end{aligned}\]
Practice
Find \(\displaystyle\int \frac{e^{2x}}{1+e^{2x}}\,dx\) and test the answer.
The denominator is the inside function. Differentiate your answer with the Chain Rule, and the result must equal the curve on top.
Solution
The denominator is the inside function, and the numerator is half of its slope.
\[u = 1 + e^{2x}, \qquad e^{2x}\,dx = \dfrac{1}{2}\,du\]Integrate in \(u\). The value of \(u\) is positive for every \(x\), so no absolute value is needed.
\[\dfrac{1}{2}\int \dfrac{du}{u} = \dfrac{1}{2}\ln u + C\]Replace \(u\). Then differentiate to test. The 2 from the Chain Rule cancels the \(\tfrac12\), which leaves the integrand.
\[\begin{aligned} &\dfrac{1}{2}\ln(1 + e^{2x}) + C \\ &\dfrac{d}{dx}\Big[\dfrac{1}{2}\ln(1 + e^{2x})\Big] \\ &\quad = \dfrac{1}{2}\cdot\dfrac{2e^{2x}}{1 + e^{2x}} \end{aligned}\]
Questions
The top curve is the integrand \(x e^{-x^2}\). The three candidates are \(A(x) = -\tfrac12 e^{-x^2}\), \(B(x) = -e^{-x^2}\) and \(C(x) = \tfrac12 e^{-x^2}\). The orange dashes show the slope that the integrand asks for.
Which candidate has a slope equal to the height of the integrand at every \(x\)?
- A
- B
- C
Answer
A
At \(x = 1\), how many times as steep as the orange tangent is the red tangent of candidate B?
- 2 times
- \(\tfrac12\) times
- The same
Answer
2 times
Which answer is correct for \(\displaystyle\int x\sqrt{1+x^2}\,dx\)?
- \(\tfrac13(1+x^2)^{3/2} + C\)
- \(\tfrac23(1+x^2)^{3/2} + C\)
- \(\tfrac13 u^{3/2} + C\)
Answer
\(\tfrac13(1+x^2)^{3/2} + C\)
Summary
You integrate in u, replace u with g(x), write one constant, and differentiate the answer to confirm it.
This works because the Chain Rule turns the answer back into the integrand, so an error in a sign or factor shows at once.
You finish a definite integral with the new limits in u or with the old limits in x, and both give the same number.
This is one of four skills in topic 6.9 of Which & Why: AP Calculus.