Class 12 Sure Board Integration Questions

This lesson works through three integration questions of the kind that often turn up on the board exam: one with an exponential factor, one solved by substitution, and one by integration by parts.

First integral: an exponential factor

We want to evaluate

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

Whenever an $e^x$ sits outside the rest, the first thing to try is the standard result

$\int \bigl(f(x) + f'(x)\bigr)\, e^x \, dx = f(x)\, e^x + C.$

That is, can we split the bracket into a function and its own derivative? Here there is no direct split, so we look at the fraction itself.

Dividing the improper fraction

The numerator and denominator both have degree $2$, so the fraction is improper. Expanding the denominator gives $(x+1)^2 = x^2 + 2x + 1$. Dividing $x^2 + 1$ by $x^2 + 2x + 1$ gives quotient $1$ and remainder $-2x$, so

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

Therefore

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

Splitting into the function plus derivative form

The remaining integral again has an $e^x$ factor, so we try the same pattern on $\dfrac{x}{(x+1)^2}$. Add and subtract $1$ in the numerator:

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

With $f(x) = \dfrac{1}{x+1}$ we have $f'(x) = -\dfrac{1}{(x+1)^2}$, so this is exactly $f(x) + f'(x)$. Hence

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

Putting it together,

$I = e^x - \frac{2}{x+1}\, e^x + C.$

Second integral: a substitution

Next,

$I = \int \frac{3ax}{b^2 + c^2 x^2}\, dx.$

Put $t = b^2 + c^2 x^2$, so $dt = 2c^2 x \, dx$ and $x \, dx = \dfrac{1}{2c^2}\, dt$. Then

$I = \frac{3a}{2c^2} \int \frac{dt}{t} = \frac{3a}{2c^2}\, \log\lvert t \rvert + C = \frac{3a}{2c^2}\, \log\bigl\lvert b^2 + c^2 x^2 \bigr\rvert + C.$

Third integral: integration by parts

Finally,

$I = \int x \, \tan x \, \sec^2 x \, dx.$

Take the first function as $x$ and the second as $\tan x \, \sec^2 x$. First we integrate the second function: with $\tan x = t$ and $\sec^2 x \, dx = dt$,

$\int \tan x \, \sec^2 x \, dx = \int t \, dt = \frac{t^2}{2} = \frac{\tan^2 x}{2}.$

Applying integration by parts,

$I = x \cdot \frac{\tan^2 x}{2} - \frac{1}{2}\int \tan^2 x \, dx.$

Using $\tan^2 x = \sec^2 x - 1$,

$\int \tan^2 x \, dx = \int (\sec^2 x - 1)\, dx = \tan x - x.$

Therefore

$I = \frac{x \, \tan^2 x}{2} - \frac{1}{2}\tan x + \frac{x}{2} + C.$

Key takeaways

• When an integrand has an $e^x$ factor, try to write the rest as $f(x) + f'(x)$, since then $\int (f + f')e^x \, dx = f e^x + C$.
• A fraction whose numerator degree is at least the denominator degree should be divided first to make it proper.
• A substitution that turns the integrand into $\tfrac{1}{t}$ produces a logarithm.
• Integration by parts works well on $x$ times a function you can integrate, and $\tan^2 x = \sec^2 x - 1$ handles the leftover integral.