If -01cot-11-x-x2dx-01tan-1x dx then - is equal to

The correct option is B 2 ∫_0^1cot^ 1(1 x+x^2)dx =∫_0^1tan^ 1 ( nbsp; x+(1 x)/1 x(1 x) )dx ∫_0^1tan^1x dx+∫_0^1tan^ 1(1 x)dx=2∫_0^1tan^ 1 ...

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

First Principle of Differentiation

If ∫01cot-1...

Question

If $\int_{0}^{1} c o t^{- 1} (1 - x + x^{2}) d x = λ \int_{0}^{1} t a n^{- 1} x d x$ then $λ$ is equal to

1

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

2

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

3

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

4

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

Open in App

Solution

Suggest Corrections

Similar questions

2 \int_{0}^{1} {tan}^{- 1} x d x = \int_{0}^{1} {cot}^{- 1} (1 - x + x^{2}) d x

\int_{0}^{1} {tan}^{- 1} (1 - x + x^{2}) d x

I = \int_{0}^{1} {cot}^{- 1} (1 - x + x^{2}) dx = k \int_{0}^{1} {tan}^{- 1} x dx

I = \int_{0}^{1} {cot}^{- 1} (1 - x + x^{2}) d x = k \int_{0}^{1} {tan}^{- 1} x d x,

k

\int_{0}^{\infty} \frac{log (1 + x^{2})}{1 + x^{2}} d x = λ \int_{0}^{1} \frac{log (1 + x)}{1 + x^{2}} d x

λ

1 \int 0 \frac{{tan}^{- 1} x}{{cot}^{- 1} (1 - x + x^{2})} d x

k,

\frac{1}{k}

Join BYJU'S Learning Program