Integration of Special Type Questions with Exponential Functions

This lesson covers a special type of integration question involving exponential functions. The key result is that whenever an integrand is an exponential times a function plus that function's derivative, the integral is just the exponential times the function. We apply it to six questions, several of which need a little rearranging first.

The standard result

The rule used throughout is:

$\int e^{x}\,\big[f(x) + f'(x)\big]\,dx = e^{x} f(x) + c$

The whole skill is recognising, or creating, this pattern: spotting a function $f(x)$ sitting next to its own derivative $f'(x)$.

Example 1: secant squared plus tangent

Evaluate:

$\int e^{x}\big(\sec^{2} x + \tan x\big)\,dx$

Reorder the bracket as $\tan x + \sec^{2} x$. Take $f(x) = \tan x$, so $f'(x) = \sec^{2} x$. The integrand is exactly $e^{x}[f(x)+f'(x)]$, so:

$\int e^{x}\big(\tan x + \sec^{2} x\big)\,dx = e^{x}\tan x + c$

Example 2: inverse tangent

Evaluate:

$\int e^{x}\left(\tan^{-1} x + \frac{1}{1+x^{2}}\right)\,dx$

Since $\dfrac{d}{dx}\big(\tan^{-1} x\big) = \dfrac{1}{1+x^{2}}$, take $f(x) = \tan^{-1} x$ and $f'(x) = \dfrac{1}{1+x^{2}}$. So:

$\int e^{x}\left(\tan^{-1} x + \frac{1}{1+x^{2}}\right)\,dx = e^{x}\tan^{-1} x + c$

Example 3: splitting a rational function

Evaluate:

$\int \frac{x^{2}+1}{(x+1)^{2}}\,e^{x}\,dx$

Here the function and its derivative are hidden, so we reshape the integrand. Write the numerator as $x^{2} - 1 + 2$ (subtract one, then add it back):

$\frac{x^{2}+1}{(x+1)^{2}} = \frac{x^{2}-1}{(x+1)^{2}} + \frac{2}{(x+1)^{2}}$

The first piece factors as $\dfrac{(x+1)(x-1)}{(x+1)^{2}} = \dfrac{x-1}{x+1}$, giving:

$\frac{x^{2}+1}{(x+1)^{2}} = \frac{x-1}{x+1} + \frac{2}{(x+1)^{2}}$

Take $f(x) = \dfrac{x-1}{x+1}$. By the quotient rule, $f'(x) = \dfrac{2}{(x+1)^{2}}$, which is the second term. So:

$\int \frac{x^{2}+1}{(x+1)^{2}}\,e^{x}\,dx = e^{x}\,\frac{x-1}{x+1} + c$

Example 4: half-angle identities

Evaluate:

$\int e^{x}\,\frac{1+\sin x}{1+\cos x}\,dx$

Split the fraction and rewrite each piece with half-angle identities, using $1+\cos x = 2\cos^{2}\tfrac{x}{2}$ and $\sin x = 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$:

$\frac{1}{1+\cos x} = \frac{1}{2}\sec^{2}\tfrac{x}{2}, \qquad \frac{\sin x}{1+\cos x} = \tan\tfrac{x}{2}$

So the integral becomes:

$\int e^{x}\left(\tan\tfrac{x}{2} + \tfrac{1}{2}\sec^{2}\tfrac{x}{2}\right)\,dx$

Take $f(x) = \tan\tfrac{x}{2}$. Then $f'(x) = \tfrac{1}{2}\sec^{2}\tfrac{x}{2}$, matching the second term. Therefore:

$\int e^{x}\,\frac{1+\sin x}{1+\cos x}\,dx = e^{x}\tan\tfrac{x}{2} + c$

Example 5: adding and subtracting in the numerator

Evaluate:

$\int \frac{x}{(1+x)^{2}}\,e^{x}\,dx$

Write the numerator $x$ as $(1+x) - 1$:

$\frac{x}{(1+x)^{2}} = \frac{1+x}{(1+x)^{2}} - \frac{1}{(1+x)^{2}} = \frac{1}{1+x} - \frac{1}{(1+x)^{2}}$

Take $f(x) = \dfrac{1}{1+x}$, so $f'(x) = -\dfrac{1}{(1+x)^{2}}$, which is the second term. Hence:

$\int \frac{x}{(1+x)^{2}}\,e^{x}\,dx = e^{x}\,\frac{1}{1+x} + c$

Example 6: double-angle identities

Evaluate:

$\int \frac{2+\sin 2x}{1+\cos 2x}\,e^{x}\,dx$

Use $\sin 2x = 2\sin x\cos x$ and $1+\cos 2x = 2\cos^{2} x$:

$\frac{2+\sin 2x}{1+\cos 2x} = \frac{2}{2\cos^{2} x} + \frac{2\sin x\cos x}{2\cos^{2} x} = \sec^{2} x + \tan x$

This is the same form as Example 1, so with $f(x) = \tan x$ and $f'(x) = \sec^{2} x$:

$\int \frac{2+\sin 2x}{1+\cos 2x}\,e^{x}\,dx = e^{x}\tan x + c$

Key takeaways

• The core rule is $\int e^{x}\big[f(x)+f'(x)\big]\,dx = e^{x} f(x) + c$.
• The work is in reshaping the integrand: split fractions, add and subtract a term, or apply trig identities until you see a function paired with its own derivative.
• Once the pattern appears, the answer is immediate: the exponential times the chosen function $f(x)$, plus the constant of integration.