Question

Let $h\left(x\right)$ be an anti derivative for $f\left(x\right)$. Then $\ln (1+(h\left(x\right){)}^3)$ is an anti derivative for

• A. $\frac{3\left(f\right(x\left){)}^{2}h\right(x)}{1+\left(h\right(x){)}^3}$
• B. $\frac{3f\left(x\right)\left(h\right(x){)}^2}{1+\left(h\right(x){)}^3}$
• C. $\frac{\left(3f\right(x\left){)}^2\right(h\left(x\right))}{1+\left(f\right(x){)}^3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{3f\left(x\right)\left(h\right(x){)}^2}{1+\left(h\right(x){)}^3}$

We have,
$h^{\prime }\left(x\right)=f\left(x\right)$ (Since $h\left(x\right)$ is the anti-derivative of $f\left(x\right)$)
Let $p\left(x\right)=\ln (1+(h\left(x\right){)}^3)$
On differentiating we get,
$p^{\prime }\left(x\right)=\frac{3h\left(x{)}^2{h}^{\prime }\right(x)}{(1+(h\left(x\right){)}^3)}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{3h\left(x{)}^{2}f\right(x)}{(1+(h\left(x\right){)}^3)}$
The anti-derivative of $p^{\prime }\left(x\right)$ is $p\left(x\right)$