Q.Integrate the following function: $\frac{6x+7}{\sqrt{(x-5)(x-4)}}$

Concept: U Substitution – we rewrite the numerator as a derivative of the quadratic inside the square root, then integrate using $\int \frac{du}{\sqrt{u}} = 2\sqrt{u}$.

First, expand the denominator: $(x-5)(x-4) = x^2 - 9x + 20$. Its derivative is $2x - 9$.

We want to express $6x+7$ as $A(2x-9) + B$.

Comparing: $6x+7 = 2Ax - 9A + B$ gives $2A = 6 \Rightarrow A = 3$, and $-9(3) + B = 7 \Rightarrow B = 34$.

So the integral becomes:

For the first term, let $u = x^2 - 9x + 20$, $du = (2x-9)dx$:

For the second term, complete the square: $x^2 - 9x + 20 = \left(x - \frac{9}{2}\right)^2 - \frac{1}{4}$. …

Expand the radicand. $(x-5)(x-4) = x^2 - 9x + 20$, whose derivative is $2x-9$.

Split the numerator. Write $6x+7 = A(2x-9) + B$. Matching coefficients gives $A = 3$ and $-9(3)+B = 7 \Rightarrow B = 34$, so $6x+7 = 3(2x-9) + 34$. Hence

First integral. With $u = x^2-9x+20$, $du = (2x-9)\,dx$:

$3\int \frac{du}{\sqrt{u}} = 6\sqrt{u} = 6\sqrt{x^2-9x+20}.$ …

Method: $\dfrac{\text{linear}}{\sqrt{\text{quadratic}}}$ split — square-root part $+$ log part

Break the linear numerator into a multiple of the radicand's derivative plus a constant; the derivative part integrates to $2\sqrt{Q}$ and the constant part to a standard root-form integral.

Steps

Step 1: Split. For $Q(x)=ax^2+bx+c$, write $px+q=\lambda(2ax+b)+\mu$ and solve for $\lambda,\mu$.

Step 2: Derivative part.

Step 3: Constant part — complete the square. …

Common Mistakes

Mistake 1: Matching the split against the wrong derivative.

Why it's wrong: the derivative of $(x-5)(x-4)=x^2-9x+20$ is $2x-9$, so $6x+7=3(2x-9)+34$; using $2x-9$ wrong drops or mis-scales the $2\sqrt{Q}$ term. Correct approach: expand the product, differentiate, then split.

Mistake 2: Forgetting the factor $2$ in the square-root part. …