Class 12 Integration: Miscellaneous Exercise 7 Solved Problems (Q28, 33, 40, 42)

This lesson solves four selected questions from the Class 12 integration miscellaneous exercise: a definite integral cleared of surds, a partial-fractions proof, an exponential integral by substitution, and an application of a definite-integral property.

Q28: A definite integral with surds

Evaluate

$I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}.$

Multiply numerator and denominator by $\sqrt{1+x} + \sqrt{x}$ to rationalise the denominator:

$I = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x}\,dx = \int_0^1 \left(\sqrt{1+x} + \sqrt{x}\right) dx,$

since the denominator simplifies to $1$.

Integrate each term:

$I = \left[\tfrac{2}{3}(1+x)^{3/2}\right]_0^1 + \left[\tfrac{2}{3}x^{3/2}\right]_0^1.$

Substituting the limits:

$I = \tfrac{2}{3}\left(2^{3/2} - 1\right) + \tfrac{2}{3}\left(1 - 0\right) = \tfrac{2}{3}\cdot 2^{3/2}.$

Since $2^{3/2} = 2\sqrt{2}$,

$I = \frac{4\sqrt{2}}{3}.$

Q33: A proof using partial fractions

Prove that

$\int_1^3 \frac{dx}{x^2(x+1)} = \frac{2}{3} + \log\frac{2}{3}.$

Partial fractions. Write

$\frac{1}{x^2(x+1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x+1},$

so that

$1 = a\,x(x+1) + b(x+1) + c\,x^2.$

Putting $x = 0$ gives $b = 1$. Putting $x = -1$ gives $c = 1$. Comparing the coefficients of $x^2$ gives $0 = a + c$, so $a = -1$.

Integrate.

$I = \int_1^3 \left(-\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}\right) dx = \left[-\log|x| - \frac{1}{x} + \log|x+1|\right]_1^3.$

Evaluating the limits:

$I = \left(-\log 3 - \tfrac{1}{3} + \log 4\right) - \left(-\log 1 - 1 + \log 2\right).$

This simplifies to

$I = \log\frac{4}{3} - \log 2 + \frac{2}{3} = \log\frac{4}{3\cdot 2} + \frac{2}{3} = \frac{2}{3} + \log\frac{2}{3},$

which is the required result.

Q40: An exponential integral by substitution

Find

$I = \int \frac{dx}{a^{x} + a^{-x}}.$

Multiply top and bottom by $a^{x}$:

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

Substitute $a^{x} = t$, so $a^{x}\ln a\,dx = dt$. This gives

$I = \frac{1}{\ln a}\int \frac{dt}{t^2 + 1} = \frac{1}{\ln a}\tan^{-1} t + c = \frac{\tan^{-1}\left(a^{x}\right)}{\ln a} + c.$

Q42: A definite-integral property

If $f(a+b-x) = f(x)$, evaluate

$I = \int_a^b x\,f(x)\,dx.$

Use the property $\int_a^b g(x)\,dx = \int_a^b g(a+b-x)\,dx$:

$I = \int_a^b (a+b-x)\,f(a+b-x)\,dx = \int_a^b (a+b-x)\,f(x)\,dx,$

using the given condition $f(a+b-x) = f(x)$.

Adding the two expressions for $I$:

$2I = \int_a^b \big[x + (a+b-x)\big] f(x)\,dx = (a+b)\int_a^b f(x)\,dx.$

Therefore

$I = \frac{a+b}{2}\int_a^b f(x)\,dx.$

Key takeaways

• Rationalising a surd denominator can turn an awkward integral into a simple sum of power functions.
• Partial fractions split a rational integrand into terms you can integrate term by term.
• A substitution like $a^{x} = t$ reduces an exponential integral to the standard form for $\tan^{-1}$.
• When $f(a+b-x) = f(x)$, the integral of $x\,f(x)$ over $[a,b]$ equals $\tfrac{a+b}{2}\int_a^b f(x)\,dx$.