Integration by Substitution: Sure Questions (Part 2)

This lesson solves four Class 12 integrals by substitution. The idea each time is to name a new variable $t$ so that the rest of the integrand turns into $dt$, leaving a much simpler integral.

Substituting $x + \log x$

We want to evaluate

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

Put $t = x + \log x$. Differentiating,

$dt = \left(1 + \frac{1}{x}\right)dx = \frac{x+1}{x}\,dx,$

which is exactly the factor sitting outside the square. So the integral becomes

$\int t^2\,dt = \frac{t^3}{3} + c.$

Substituting back,

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

Top is the derivative of the bottom

Next,

$\int \frac{10x^{9} + 10^{x}\log 10}{x^{10} + 10^{x}}\,dx.$

Put $t = x^{10} + 10^{x}$. Using $\tfrac{d}{dx}\,a^{x} = a^{x}\log a$,

$dt = \left(10x^{9} + 10^{x}\log 10\right)dx,$

so the numerator is precisely $dt$. Calling the integral $I$,

$I = \int \frac{dt}{t} = \log|t| = \log\bigl(x^{10} + 10^{x}\bigr) + c.$

Since $x^{10} + 10^{x}$ is positive, the modulus is not needed and we can drop it.

Special case: adjusting the numerator

This one needs care:

$\int \frac{x}{\sqrt{x+4}}\,dx.$

If we simply put $x+4 = t$, the numerator $x$ does not appear as $dt$. The trick is to write the numerator as $x+4$ by adding and subtracting $4$:

$\int \frac{x}{\sqrt{x+4}}\,dx = \int \frac{(x+4) - 4}{\sqrt{x+4}}\,dx = \int \sqrt{x+4}\,dx - 4\int \frac{1}{\sqrt{x+4}}\,dx.$

Integrating each piece,

$\int \sqrt{x+4}\,dx = \frac{(x+4)^{3/2}}{3/2} = \frac{2}{3}(x+4)^{3/2},$

$4\int \frac{1}{\sqrt{x+4}}\,dx = 4 \cdot 2\sqrt{x+4} = 8\sqrt{x+4}.$

Therefore

$\int \frac{x}{\sqrt{x+4}}\,dx = \frac{2}{3}(x+4)^{3/2} - 8\sqrt{x+4} + c.$

Splitting a higher power with substitution

Finally,

$\int x^{5}\,(x^{3}-1)^{1/3}\,dx.$

We split $x^{5}$ as $x^{3}\cdot x^{2}$ so that one factor matches the substitution. Put $t = x^{3} - 1$, so $x^{3} = t + 1$ and

$x^{2}\,dx = \tfrac{1}{3}\,dt.$

Then $x^{5}\,dx = x^{3}\cdot x^{2}\,dx = (t+1)\cdot \tfrac{1}{3}\,dt$, and the integral becomes

$\frac{1}{3}\int t^{1/3}(t+1)\,dt = \frac{1}{3}\int \left(t^{4/3} + t^{1/3}\right)dt.$

Integrating,

$\frac{1}{3}\left(\frac{t^{7/3}}{7/3} + \frac{t^{4/3}}{4/3}\right) = \frac{1}{7}\,t^{7/3} + \frac{1}{4}\,t^{4/3}.$

Substituting $t = x^{3}-1$ back,

$\int x^{5}\,(x^{3}-1)^{1/3}\,dx = \frac{1}{7}(x^{3}-1)^{7/3} + \frac{1}{4}(x^{3}-1)^{4/3} + c.$

Key takeaways

• Choose the substitution so the leftover part of the integrand becomes $dt$.
• When the numerator is the exact derivative of the denominator, the integral is $\log$ of the denominator.
• If a substitution does not give the term you need, add and subtract a constant in the numerator and split the integral.
• For a high power like $x^{5}$, peel off the factor that matches $dt$ and express the rest through the new variable.