The value of -d x-xn1-xn1-n- n - NA- 1-1-n1-1-xn1-1-n- cB- 1-1- n 1-1- x n 1-1- n - cC- -1-1-n1-1-xn1-1-n- cD- 1-1- n 1-1- x n 1-1- n - c

The correct option is D 1/1 n{ 1+1/x^n}^1 1/n+cI=∫dx/x^n(1+x^n)^1/n=∫dx/x^n+1 ( 1+1/x^n )^1/n Let 1+1/x^n=t⇒ n/x^n+1dx= dt ∴ I= 1/n∫dt/t^1/n = 1/n [ t^ 1/n+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration by Substitution

The value of ...

Question

The value of $\int \frac{d x}{x^{n} (1 + x^{n})^{\frac{1}{n}}}, n ϵ N$

\frac{1}{1 - n} {1 + \frac{1}{x^{n}}}^{1 - \frac{1}{n}} + c

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

\frac{1}{1 + n} {1 - \frac{1}{x^{n}}}^{1 - \frac{1}{n}} + c

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

- \frac{1}{1 - n} {1 - \frac{1}{x^{n}}}^{1 - \frac{1}{n}} + c

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

\frac{1}{1 - n} {1 + \frac{1}{x^{n}}}^{1 - \frac{1}{n}} + c

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

Open in App

Solution

The correct option is D $\frac{1}{1 - n} {1 + \frac{1}{x^{n}}}^{1 - \frac{1}{n}} + c$
$I = \int \frac{d x}{x^{n} (1 + x^{n})^{1 / n}} = \int \frac{d x}{x^{n + 1} {(1 + \frac{1}{x^{n}})}^{1 / n}} L e t 1 + \frac{1}{x^{n}} = t \Rightarrow \frac{- n}{x^{n + 1}} d x = d t ∴ I = - \frac{1}{n} \int \frac{d t}{t^{\frac{1}{n}}} = - \frac{1}{n ⎡ ⎣ \frac{t^{- \frac{1}{n} + 1}}{- \frac{1}{n} + 1} ⎤ ⎦} + c = - \frac{{(1 + \frac{1}{x^{n}})}^{1 - \frac{1}{n}}}{n - 1} + c$

Suggest Corrections

Similar questions

Q. Assertion :If

n > 1

then

\int_{0}^{\infty} \frac{d x}{1 + x^{n}} = \int_{0}^{1} \frac{d x}{{(1 - x^{n})}^{1 / n}}

Reason:

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

Q. Assertion :If

n > 1

then Statement -1:

\int_{0}^{\infty} \frac{d x}{1 + x^{n}} = \int_{0}^{1} \frac{d x}{(1 - x^{n})^{1 / n}}

Reason: Statement -2:

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

Q. The coefficient of

\frac{1}{x}

in the expansion of

(1 + x)^{n} {(1 + \frac{1}{x})}^{n}

Q. The coefficient of

x^{- n}

(1 + x)^{n} {(1 + \frac{1}{x})}^{n}

Q. The coefficient of

\frac{1}{x}

in the expansion of

{(1 + x)}^{n} {(1 + \frac{1}{x})}^{n}

is
(a)

\frac{n!}{[(n - 1)! (n + 1)!]}

(b)

\frac{(2 n)!}{[(n - 1)! (n + 1)!]}

(c)

\frac{(2 n)!}{(2 n - 1)! (2 n + 1)!}

(d) none of these

Join BYJU'S Learning Program

The value of ∫dxxn(1+xn)1n,nϵN

The correct option is D 11−n{1+1xn}1−1n+cI=∫dxxn(1+xn)1/n=∫dxxn+1(1+1xn)1/nLet 1+1xn=t⇒−nxn+1dx=dt∴I=−1n∫dtt1n=−1n⎡⎣t−1n+1−1n+1⎤⎦+c=−(1+1xn)1−1nn−1+c

The value of $\int \frac{d x}{x^{n} (1 + x^{n})^{\frac{1}{n}}}, n ϵ N$