If $\int e^{2x}f' \left(x\right)dx =g \left(x\right)$, then $ \int\left(e^{2x}f\left(x\right) + e^{2x} f' \left(x\right)\right)dx =$
- $\frac{1}{2} [e^{2x} f(x) - g (x)] + C $
- $\frac{1}{2} [e^{2x} f(x) + g (x)] + C $
- $\frac{1}{2} [e^{2x} f'(2x) + g (x)] + C $
- $\frac{1}{2} [e^{2x} f'(2x) + g (x)] + C $
The Correct Option is A
Solution and Explanation
$\int e^{2 x} f^{\prime}(x) d x=g(x)$
Let $I=\int\left(e^{2 x} f(x)+e^{2 x} f^{\prime}(x)\right) d x$
$\left.=f(x) \int e^{2 x} d x-\int f^{\prime}(x) \int e^{2 x} d x\right) d x+\int e^{2 x} f^{\prime}(x) d x$
$=\frac{f(x) e^{2 x}}{2}-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x+\int e^{2 x} f^{\prime}(x) d x$
$=\frac{e^{2 x}}{2} f(x)-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x$
$=\frac{1}{2}\left[e^{2 x} f(x)-\int e^{2 x} f^{\prime}(x) d x\right]$
$=\frac{1}{2}\left[e^{2 x} f(x)-g(x)\right]+C$
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C