AHL 5.11—Indefinite integration, reverse chain, by substitution

The integrand is $(ax+b{)}^n$ with $a=4$, $b=5$, $n=3$.

$\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C$.

$\frac{(4x+5{)}^4}{4(4)}+C=\frac{(4x+5{)}^4}{16}+C$.

Differentiating gives $\frac{4(4x+5{)}^3\cdot 4}{16}=(4x+5{)}^3$, confirming the result.

$\frac{(4x+5{)}^4}{4}$ divides only by $a$, forgetting to also divide by $n+1=4$.

$(4x+5{)}^4$ ignores the need to divide by $a(n+1)$ entirely, as if $a=n+1=1$.

$\frac{(4x+5{)}^4}{12}$ uses $n+1=3$ instead of $4$ in the denominator's second factor, giving $a\cdot n=12$.

Only option A correctly divides by $a(n+1)=16$.

This is $\int \frac{1}{ax+b}dx$ with $a=2$, $b=-7$.

$\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+C$.

$\frac{1}{2}\ln |2x-7|+C$.

Differentiating gives $\frac{1}{2}\cdot \frac{2}{2x-7}=\frac{1}{2x-7}$, matching the integrand.

Forgetting the $\frac{1}{a}$ factor entirely gives $\ln |2x-7|+C$, which differentiates to $\frac{2}{2x-7}$, not $\frac{1}{2x-7}$.

Multiplying by $a$ instead of dividing gives $2\ln |2x-7|+C$, the reciprocal mistake.

$\frac{1}{2}\ln |x|+C$ drops the linear expression inside the log entirely.

Only B correctly divides by $a=2$ and keeps $2x-7$ inside the logarithm.