displaystyle ∫ (xn-1/x2n+a2) dx dX is equal to

Question

displaystyle ∫ (xn-1/x2n+a2) dx dX is equal to (A) (1/na) tan-1 ((xn/a)) +C (B) (n/a) tan-1 ((xn/a)) +C (C) (n/a) sin-1 ((xn/a)) +C (D) (n/a) cos-1 ((xn/a)) +C

Tardigrade Admin · Accepted Answer

Let I=∫ (xn-1/x2 n+a2) d x operatornamePut xn=t ⇒ n xn-1 d x=d t ∴ I=(1/n) ∫ (d t/t2+a2)=(1/n) ⋅ (1/a) tan -1 (t/a)+C Put t=xn ⇒ I=(1/n a) tan -1((Xn/a))+C

Q. $\int \frac{x ^{n - 1}}{x ^{2 n} + a ^{2}} d x$ dX is equal to

$\frac{1}{na} t a n^{- 1} (\frac{x ^{n}}{a}) + C$

$\frac{n}{a} t a n^{- 1} (\frac{x ^{n}}{a}) + C$

$\frac{n}{a} s i n^{- 1} (\frac{x ^{n}}{a}) + C$

$\frac{n}{a} co s^{- 1} (\frac{x ^{n}}{a}) + C$

$\frac{1}{na} co t^{- 1} (\frac{x ^{n}}{a}) + C$

Let $I = \int \frac{x ^{n - 1}}{x ^{2 n} + a ^{2}} d x$
$Put x^{n} = t$
$\Rightarrow n x^{n - 1} d x = d t$
$∴ I = \frac{1}{n} \int \frac{d t}{t ^{2} + a ^{2}} = \frac{1}{n} \cdot \frac{1}{a} tan^{- 1} \frac{t}{a} + C$
Put $t = x^{n}$
$\Rightarrow I = \frac{1}{na} tan^{- 1} (\frac{X ^{n}}{a}) + C$

Q. ∫x2n+a2xn−1​dxdX is equal to

na1​tan−1(axn​)+C

an​tan−1(axn​)+C

an​sin−1(axn​)+C

an​cos−1(axn​)+C

na1​cot−1(axn​)+C