Q30 / 50 · Manipal (MET) 2022 · Mathematics
If $\int f(x) d x=g(x)$, then $\int x^9 \cdot f\left(x^5\right) d x$ is equal to
- A$\frac{1}{5}\left(x^5 g\left(x^9\right)-4 \int g(x) d x+C\right.$
- B$\frac{1}{5}\left[x^9 g\left(x^5\right)-\frac{1}{5} \int x^4 g\left(x^5\right) d x+C\right.$
- C$\frac{1}{5}\left[g\left(x^9\right)+\int g\left(x^5\right) d x\right]$
- Correct answerD$\frac{x^5}{5} g\left(x^5\right)-\int x^4 g\left(x^5\right) d x+C$
Solution
Given that, $\int f(x) d x=g(x)$
Let $\quad I=\int x^9 \int\left(x^5\right) d x$ Put $x^5=t \Rightarrow 5 x^4 d x=d t$ $\begin{aligned} & =\frac{1}{5} \int t f(t) d t \\ & =\frac{1}{5}\left[t \int f(t) d t-\int[f(t) d t]\right. \end{aligned}$ [integrating by part] $\begin{aligned} & =\frac{1}{5}\left[t g(t)-\int g(t) d t\right] \\ & =\frac{1}{5}\left[x^2 g\left(x^5\right)-5 \int g\left(x^5\right) x^4 d x\right]+C \\ & =\frac{x^5}{5} g\left(x^5\right)-\int x^4 g\left(x^5\right) d x+C \end{aligned}$
Let $\quad I=\int x^9 \int\left(x^5\right) d x$ Put $x^5=t \Rightarrow 5 x^4 d x=d t$ $\begin{aligned} & =\frac{1}{5} \int t f(t) d t \\ & =\frac{1}{5}\left[t \int f(t) d t-\int[f(t) d t]\right. \end{aligned}$ [integrating by part] $\begin{aligned} & =\frac{1}{5}\left[t g(t)-\int g(t) d t\right] \\ & =\frac{1}{5}\left[x^2 g\left(x^5\right)-5 \int g\left(x^5\right) x^4 d x\right]+C \\ & =\frac{x^5}{5} g\left(x^5\right)-\int x^4 g\left(x^5\right) d x+C \end{aligned}$
Hence, the correct answer is option: 'D'.