H2 - Standard exponential and trigonometric integrals | AQA A-Level Mathematics

Integrate exponentials, reciprocals, sine and cosine by reversing their derivatives. Learn how the coefficient inside an argument changes the answer, why logarithms need a valid domain and how to combine standard forms.

Exponential integrals and scale factors

Which function has derivative $e^{2x}$ : $2e^{2x}$ or $\frac12e^{2x}$ ? Differentiating the first gives $4e^{2x}$ ; differentiating the second gives exactly $e^{2x}$ . Integration must undo the factor introduced by the chain rule.

You already know

$\frac{d}{dx}e^{kx+b}=ke^{kx+b}$ . Also $a^u=e^{u\ln a}$ for $a>0$ , so $\frac{d}{dx}a^{kx+b}=k\ln(a)\,a^{kx+b}$ . These derivative and logarithm results are not in the formula booklet: learn them.

The coefficient $k$ is the derivative of the linear argument. The shift $b$ contributes no further factor because its derivative is zero.

Exponential antiderivatives

For constants $A$ , $b$ and $k\ne0$ ,

\int Ae^{kx+b}\,dx=\frac Ak e^{kx+b}+C

For a constant base $a>0$ , $a\ne1$ ,

\int Aa^{kx+b}\,dx=\frac{A}{k\ln a}a^{kx+b}+C

Both hold for every real $x$ . If $k=0$ , or if $a=1$ , integrate the resulting constant directly.

Not in the formula booklet: learn it.

This is the reverse chain rule for a linear argument: divide by the derivative of the inside. For example, $\int5e^{2x+1}\,dx=\frac52e^{2x+1}+C$ ; differentiating multiplies $5/2$ by $2$ and recovers $5$ .

Worked example 1: two exponential rates

Question. Find $\displaystyle\int(4e^{2x}-3e^{-x})\,dx$ .

Plan. Divide each coefficient by its exponential's inner coefficient.

\begin{aligned} I&=\frac42e^{2x}-\frac3{-1}e^{-x}+C && \text{divide by }2\text{ and }-1 \\ &=2e^{2x}+3e^{-x}+C && \text{simplify signs} \end{aligned}

Why this step? The negative inner coefficient reverses the sign of the second antiderivative term.

Answer. $2e^{2x}+3e^{-x}+C$ .

Check. Its derivative is $4e^{2x}-3e^{-x}$ .

Change only the first exponential's base from $e$ to $2$ . The extra $\ln2$ comes from its derivative, so it must be divided out too.

Worked example 2: a different base

Question. Find $\displaystyle\int(4\cdot2^{2x}-3e^{-x})\,dx$ .

Plan. Use the base- $2$ formula for the first term and the same second term as before.

\begin{aligned} I&=\frac4{2\ln2}2^{2x}+3e^{-x}+C && \text{divide by }2\ln2 \\ &=\frac2{\ln2}2^{2x}+3e^{-x}+C && \text{simplify} \end{aligned}

Why this step? Replacing the base changes the derivative's multiplier, even though the exponent still reads $2x$ .

Answer. $\frac2{\ln2}2^{2x}+3e^{-x}+C$ .

Check. $\frac2{\ln2}(2\ln2)2^{2x}-3e^{-x}=4\cdot2^{2x}-3e^{-x}$ .

Differentiation gives a quick way to test every coefficient.

Watch out

Wrong: $\int e^{3x}\,dx=3e^{3x}+C$ . Its derivative is $9e^{3x}$ . Right: $\frac13e^{3x}+C$ , because $\frac13\times3=1$ .

The next integral changes just the first base again. Keep the logarithm exact.

Reciprocal integrals and logarithm domains

The reciprocal $1/x$ is the power-rule exception. For $x>0$ , it is the derivative of $\ln x$ . For $x<0$ , $\ln(-x)$ has derivative $(-1)/(-x)=1/x$ . Writing $\ln|x|$ covers either interval with a positive logarithm argument.

The supplied logarithmic rule

For differentiable $f$ , on an interval where $f(x)\ne0$ ,

\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C

In the formula booklet.

For a linear denominator $ax+b$ , its derivative is $a$ . Multiplying the supplied rule by $A/a$ gives the useful special case below.

Reciprocal antiderivatives

\int\frac1x\,dx=\ln|x|+C \qquad (x\ne0)

For $a\ne0$ , on an interval avoiding $ax+b=0$ ,

\int\frac{A}{ax+b}\,dx=\frac Aa\ln|ax+b|+C

These special cases can be recovered from the supplied logarithmic rule.

Not in the formula booklet: learn it.

The absolute value does not fill the hole at zero. Constants may differ on disconnected intervals: an antiderivative of $1/x$ can be $\ln(-x)+C_-$ for $x<0$ and $\ln x+C_+$ for $x>0$ , with no requirement that $C_-=C_+$ . A definite integral across zero is not justified by subtracting these endpoint values. On a stated positive-argument domain, the absolute value can be omitted; AQA does not require modulus notation in those standard positive-domain answers.

Worked example 3: a linear denominator

Question. Find $\displaystyle\int\frac3{2x-1}\,dx$ on an interval avoiding $x=1/2$ .

Plan. Divide the numerator coefficient by the denominator's derivative, $2$ .

\begin{aligned} I&=\frac32\int\frac2{2x-1}\,dx && \text{match numerator to derivative} \\ &=\frac32\ln|2x-1|+C && \text{use the log rule} \end{aligned}

Why this step? Differentiating the logarithm introduces $2$ ; the coefficient $3/2$ restores the required numerator $3$ .

Answer. $\frac32\ln|2x-1|+C$ , on either side of $x=1/2$ .

Check. Differentiation gives $(3/2)\cdot2/(2x-1)=3/(2x-1)$ . For $x<1/2$ , the answer is also $\frac32\ln(1-2x)+C$ .

Change only the numerator from $3$ to $5$ . The excluded point and logarithm argument stay the same.

Worked example 4: a new numerator

Question. Find $\displaystyle\int\frac5{2x-1}\,dx$ on an interval avoiding $x=1/2$ .

Plan. Match the denominator's derivative and keep the remaining constant outside.

\begin{aligned} I&=\frac52\int\frac2{2x-1}\,dx && \text{factor out }5/2 \\ &=\frac52\ln|2x-1|+C && \text{integrate} \end{aligned}

Why this step? A constant multiplier scales the entire antiderivative; it does not go inside the logarithm as a multiplier of $x$ .

Answer. $\frac52\ln|2x-1|+C$ .

Check. $\frac{d}{dx}\left(\frac52\ln|2x-1|\right)=5/(2x-1)$ wherever $2x-1\ne0$ .

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H2 - Standard exponential and trigonometric integrals

Exponential integrals and scale factors

Reciprocal integrals and logarithm domains

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