Indefinite Integration
Indefinite Integration
nta_pyq_2025_apr
Grade 12
Question:
If $f(x) = \displaystyle\int \frac{1}{x^{1/4}(1+x^{1/4})}\,dx$, $f(0) = -6$, then $f(1)$ is equal to:
$4(\log_e 2 - 2)$
$2 - \log_{e^2}\!2$
$\log_e 2 + 2$
$4(\log_e 2 + 2)$
Step-by-Step Solution
Key Concept: Substitute $t = x^{1/4}$ so $dx = 4t^3\,dt$ and the integrand becomes $\dfrac{4t^3}{t(1+t)} = 4\cdot\dfrac{t^2}{1+t}$; perform polynomial division to split into manageable terms.
Put $x^{1/4} = t$, $dx = 4t^3\,dt$:
$$\int\frac{4t^3}{t(1+t)}dt = 4\int\frac{t^2}{1+t}dt = 4\int\!\left(t-1+\frac{1}{1+t}\right)dt.$$
$$= 4\left[\frac{t^2}{2} - t + \ln|1+t|\right] + C = 4\left[\frac{x^{1/2}}{2} - x^{1/4} + \ln(1+x^{1/4})\right]+C.$$
$f(0) = 0 + C = -6 \Rightarrow C = -6$.
$$f(1) = 4\!\left(\frac{1}{2}-1+\ln 2\right)-6 = 4\ln 2 - 2 - 6 = 4(\ln 2 - 2).$$
Correct Answer: 1