Q.Find the following integrals:

Concept: U Substitution — rewrite the numerator as a multiple of the derivative of the denominator (or the expression inside the square root) plus a constant, then split into two standard integrals.

(i) $\int \dfrac{x + 2}{2x^2 + 6x + 5}\, dx$

Derivative of denominator: $4x + 6 = 2(2x + 3)$. Write numerator as $x+2 = \frac{1}{4}(4x+6) + \frac{1}{2}$.

Then the integral becomes:

First part: $u = 2x^2+6x+5$, $du = (4x+6)dx$, gives $\frac{1}{4}\log|2x^2+6x+5|$. …

(i) $\displaystyle \int \frac{x + 2}{2x^2 + 6x + 5}\, dx$

Concept and intuition

When the denominator is a quadratic, the derivative of the denominator is $4x + 6$. The numerator $x+2$ is almost a multiple of $4x+6$, but not quite. The trick is to write the numerator as:

where $A$ and $B$ are constants. This splits the integral into two pieces:

• The part with $A$ gives $\int \frac{4x+6}{2x^2+6x+5}\,dx$, which is a simple logarithm (since the numerator is exactly the derivative of the denominator).
• The part with $B$ gives $\int \frac{1}{2x^2+6x+5}\,dx$, which after completing the square becomes an inverse tangent.

Step‑by‑step

1. Find $A$ and $B$.

We want $x+2 = A(4x+6) + B$.

Comparing coefficients:

Coefficient of $x$: $1 = 4A \implies A = \frac14$.

Constant term: $2 = 6A + B \implies 2 = 6\cdot\frac14 + B = \frac32 + B \implies B = \frac12$.

So:

2. Split the integral.

3. First integral — the log part.

Let $u = 2x^2+6x+5$, then $du = (4x+6)\,dx$. So:

4. Second integral — prepare for $\tan^{-1}$.

Factor the $2$ from the denominator:

Complete the square:

So the integral becomes:

5. Use the standard form.

Recall $\displaystyle \int \frac{1}{t^2 + a^2}\,dt = \frac1a \tan^{-1}\!\left(\frac{t}{a}\right) + C$.

Here $t = x+\frac32$, $a = \frac12$. So:

6. Combine the results.

Method: Split a Linear Numerator into (Derivative of Denominator) + Constant

Use this when a linear numerator sits over a quadratic (or its square root): rewrite the numerator so one part is proportional to the denominator's derivative, splitting the integral into a log/root part and an arctan/arcsine part.

Steps

Step 1: Write the numerator as $A\cdot(\text{derivative of denominator})+B$.

If the denominator is $q(x)$, set numerator $=A\,q'(x)+B$ and match coefficients to find $A$ and $B$. E.g. for $2x^2+6x+5$, $q'(x)=4x+6$, and $x+2=A(4x+6)+B$.

Step 2: Split the integral into two standard pieces. …

Common Mistakes

Mistake 1: Guessing $A$ and $B$ instead of solving.

Why it's wrong: the split must satisfy $x+2=A(4x+6)+B$ exactly; a guess leaves a mismatch. Correct approach: equate coefficients of $x$ and the constant to solve for $A$ and $B$.

Mistake 2: Forgetting to complete the square on the constant-numerator piece. …