Question

$\int f^{\prime }(ax+b){\left\{f\right(ax+b\left)\right\}}^{n}dx$ is equal to

• A. $\frac{1}{n+1}{\left\{f\right(ax+b\left)\right\}}^{n+1}+C$, for all $n$ except $n=-1$
• B. $\frac{1}{n+1}{\left\{f\right(ax+b\left)\right\}}^{n+1}+C$, for all $n$
• C. $\frac{1}{a(n+1)}{\left\{f\right(ax+b\left)\right\}}^{n+1}+C$, for all $n$ except $n=-1$
• D. $\frac{1}{a(n+1)}{\left\{f\right(ax+b\left)\right\}}^{n+1}+C$, for all $n$

Answer 1

The correct option is C $\frac{1}{a(n+1)}{\left\{f\right(ax+b\left)\right\}}^{n+1}+C$, for all $n$ except $n=-1$

$\int f^{\prime }(ax+b)\left(f\right(ax+b){)}^{n}dx$
$f(ax+b)\to t$
$a{f}^1(ax+b)\to \frac{dt}{dx}$
$f^1(ax+b)dx\to \frac{dt}{a}$
Therefore, $e\int \frac{t^n\cdot dt}{a}=\frac{t^{n+1}}{a(n+1)}+c$
$=\frac{\left(f\right(ax+b){)}^{n+1}}{a(n+1)}+c$