TS EAMCET · Maths · Indefinite Integration
If \(\int \phi(x) d x=\psi(x)\), then \(\int\left(\phi_0 h\right)(x) h(x) h^{\prime}(x) d x=\)
- A \((\phi h)(x) \phi^{\prime}(x)-\int(\phi h)(x) h^{\prime}(x) d x+c\)
- B \(\left(\psi_0 h\right)(x) h(x)-\int\left(\psi_0 h\right)(x) h^{\prime}(x) d x+c\)
- C \(\left(\psi_0 h\right)(x) \phi(x)-\int\left(\psi_0 h\right)(x) \phi^{\prime}(x) d x+c\)
- D \(\left(\psi_0 \phi\right)(x) h(x)-\int\left(\psi_0 \phi\right)(x) h^{\prime}(x) d x+c\)
Answer & Solution
Correct Answer
(B) \(\left(\psi_0 h\right)(x) h(x)-\int\left(\psi_0 h\right)(x) h^{\prime}(x) d x+c\)
Step-by-step Solution
Detailed explanation
\(\int \phi(x) d x=\psi(x)\), then \(\int(\phi \circ h)(x) \cdot h(x) h^{\prime}(x) d x=\) ? Now, \(\int \phi\{h(x)\} \cdot h(x) \cdot h^{\prime}(x) d x\) Let \(h(x)=t \Rightarrow h^{\prime}(x) d x=d t\) Now, \(\int_{\mathrm{II}}^{\phi(t) \cdot{ }_{\mathrm{I}}} d t\)…
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