If u and v are two functions of x- then prove that - uv dx - u - v dx - - -dudx- v dx - dx

Let u and v be any two functions of x . #160;Then, by product rule of differentiation ddx(u v) = u dvdx + vdudx Differentiating both side ...

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Standard XII

Mathematics

Integration by Parts

If u and ...

Question

If $u$ and $v$ are two functions of $x$ , then prove that
$\int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$

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Solution

Let $u$ and $v$ be any two functions of $x$ .
Then, by product rule of differentiation
$\frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x}$
Integrating both sides with respect to $x$ ,
$u v = \int u \frac{d v}{d x} d x + \int v \frac{d u}{d x} d x$
$\Rightarrow \int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$
Now, put $u = f_{1} (x)$ and $\frac{d v}{d x} = f_{2} (x)$ , i.e., $v = \int f_{2} (x) d x$
$\int f_{1} (x) f_{2} (x) d x = f_{1} (x) \int f_{2} (x) d x - \int [\frac{d}{d x} {f_{1} (x)} \int f_{2} (x) d x] d x$
The above expression can be rewrite as
$\Rightarrow \int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$
Where, $u = f_{1} (x), v = f_{2} (x)$ .

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If u and v are two functions of x, then prove that∫uv dx=u∫v dx−∫[dudx∫v dx]dx

If $u$ and $v$ are two functions of $x$ , then prove that
$\int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$