Integration: Simple Questions (Part 1)

This lesson works through five simple indefinite integrals. In each one we first reshape the integrand into terms we can integrate directly, then apply the basic power and exponential rules and add the constant of integration.

Exponential plus a constant

Evaluate $\int \left(4e^{3x} + 1\right)\, dx$.

Split the integral into two terms:

$\int 4e^{3x}\, dx + \int 1\, dx.$

The integral of $e^{3x}$ is $\tfrac{1}{3}e^{3x}$ (we divide by the derivative of $3x$, which is $3$), and the integral of $1$ is $x$. So

$\int \left(4e^{3x} + 1\right)\, dx = \frac{4}{3}e^{3x} + x + C.$

Expanding a squared bracket

Evaluate $\int \left(\sqrt{x} - \tfrac{1}{\sqrt{x}}\right)^2 dx$.

Expand using $(a-b)^2 = a^2 - 2ab + b^2$:

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

The middle term simplifies because $\sqrt{x}\cdot\tfrac{1}{\sqrt{x}} = 1$. Now integrate term by term:

$\int \left(x - 2 + \frac{1}{x}\right) dx = \frac{x^2}{2} - 2x + \ln|x| + C.$

Dividing each term by the denominator

Evaluate $\displaystyle\int \frac{x^3 + 5x^2 - 4}{x^2}\, dx$.

Distribute the $x^2$ across each term in the numerator:

$\int \left(x + 5 - \frac{4}{x^2}\right) dx.$

Writing $\tfrac{1}{x^2} = x^{-2}$, its integral is $\tfrac{x^{-1}}{-1} = -\tfrac{1}{x}$. Therefore

$\int \frac{x^3 + 5x^2 - 4}{x^2}\, dx = \frac{x^2}{2} + 5x + \frac{4}{x} + C.$

Roots as fractional powers

Evaluate $\displaystyle\int \frac{x^2 + 3x + 4}{\sqrt{x}}\, dx$.

Since $\sqrt{x} = x^{1/2}$, dividing each term lowers its power by $\tfrac{1}{2}$:

$\int \left(x^{3/2} + 3x^{1/2} + 4x^{-1/2}\right) dx.$

Apply the power rule to each term:

$\frac{x^{5/2}}{5/2} + 3\cdot\frac{x^{3/2}}{3/2} + 4\cdot\frac{x^{1/2}}{1/2} = \frac{2}{5}x^{5/2} + 2x^{3/2} + 8\sqrt{x} + C.$

Factorise and cancel before integrating

Evaluate $\displaystyle\int \frac{x^3 - x^2 + x - 1}{x - 1}\, dx$.

The denominator is a single linear factor, so we cannot split the fraction term by term. Instead factorise the numerator by grouping:

$x^3 - x^2 + x - 1 = x^2(x - 1) + (x - 1) = (x - 1)(x^2 + 1).$

Cancel the common factor $x - 1$:

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

Key takeaways

• Reshape the integrand first: split sums, expand squares, divide through, or convert roots to fractional powers so each term integrates directly.
• The integral of $e^{ax}$ is $\tfrac{1}{a}e^{ax}$, and $\int \tfrac{1}{x}\, dx = \ln|x|$.
• When a linear denominator blocks splitting, factorise the numerator and cancel before integrating.