$\int x (f (x^{2}) g^{''} (x^{2}) - f^{''} (x^{2}) g (x^{2})) d x =$

f (x^{2}) g^{'} (x^{2}) - g (x^{2}) f^{'} (x^{2}) + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} (f (x^{2}) g (x^{2}) f^{'} (x^{2})) + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} (f (x^{2}) g^{'} (x^{2}) - g (x^{2}) f^{'} (x^{2})) + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $\frac{1}{2} (f (x^{2}) g^{'} (x^{2}) - g (x^{2}) f^{'} (x^{2})) + c$
$I = \int x (f (x^{2}) g^{''} (x^{2}) - f^{''} (x^{2}) g (x^{2})) d x$
Substitute $x^{2} = t \Rightarrow 2 x d x = d t$
$I = \frac{1}{2} \int (f (t) g^{''} (t) - g (t) f^{''} (t)) d t$
$= \frac{1}{2} \int f (t) g^{''} (t) d t - \frac{1}{2} \int g (t) f^{''} (t)$
$= \frac{1}{2} (f (t) g^{'} (t) - \int f^{'} (t) g^{'} (t) d t - g (t) f^{'} (t) + \int g^{'} (t) f^{'} (t) d t) + c$
$= \frac{1}{2} (f (t) g^{'} (t) - g (t) f^{'} (t)) + c$
$= \frac{1}{2} (f (x^{2}) g^{'} (x^{2}) - g (x^{2}) f^{'} (x^{2})) + c$

Suggest Corrections

Similar questions

\int x {f (x^{2}) g^{''} (x^{2}) - f^{''} (x^{2}) g (x^{2})} d x

If $α, β$ are roots of the equation $x^{2} - p (x + 1) - c = 0$ , then

$(α + 1) (β + 1)$ =

Q. If

x ϵ R

, then

\frac{x^{2} - x + 1}{x^{2} + x + 1}

takes values in the interval

Q. sin

[\cot^{- 1} \{\tan (\cos^{- 1} x)\}]

is equal to
(a) x

(b)

\sqrt{1 - x^{2}}

(c)

\frac{1}{x}

(d) none of these

If f (x) is differentiable in the interval [2, 5], where $f (2) = \frac{1}{5}$ and $f (5) = \frac{1}{2}$ , then there exists a number c, 2 < c < 5 for which f ' (c) is equal to

Join BYJU'S Learning Program

∫x(f(x2)g′′(x2)−f′′(x2)g(x2))dx=

$\int x (f (x^{2}) g^{''} (x^{2}) - f^{''} (x^{2}) g (x^{2})) d x =$