-01 e tan -1 x -1- x 2 dx is equal toA- - e -4B- 1 C- e-4-1D- e-4-1

The correct option is D e^π/4 1∫_0^1e^tan^ 1x1+x^2 dx Let nbsp; tan^ 1 x =t ⇒11+x^2dx=dt =∫_0^π/4e^t dt=e^t|_0^π/4 =e^π/4 1 ...

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

Byju's Answer

Standard XII

Mathematics

Integration of Irrational Algebraic Fractions - 2

∫01 e tan -1 ...

Question

$1 \int 0 \frac{e^{{tan}^{- 1} x}}{1 + x^{2}} d x$ is equal to

- e^{\frac{π}{4}}

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

1

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

e^{\frac{π}{4}} + 1

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

e^{\frac{π}{4}} - 1

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

Open in App

Solution

The correct option is D $e^{\frac{π}{4}} - 1$
$1 \int 0 \frac{e^{{tan}^{- 1} x}}{1 + x^{2}} d x$
Let ${tan}^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} d x = d t$
$= \frac{π}{4} \int 0 e^{t} d t = e^{t} {∣ ∣}_{0}^{\frac{π}{4}} = e^{\frac{π}{4}} - 1$

Suggest Corrections

Similar questions

Q. The integral

\frac{π}{2} \int \frac{π}{4} e^{x} (ln (cos x)) (cos x sin x - 1) {cosec}^{2} x d x

is equal to

Q. The value of

\frac{π}{2} \int - \frac{π}{2} \frac{x^{2} cos x}{1 + e^{x}} d x

is equal to

The option with the values of L that satisfying the equation $\frac{\int_{0}^{4 π} e^{t} (s i n^{6} a t + c o s^{4} a t) d t}{\int_{0}^{π} e^{t} (s i n^{6} a t + c o s^{4} a t) d t} = L,$ is

If z = $i^{i}$ . Express ${l o g}_{e}$ z in A+iB form.Find the value of A and B.

\int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (l o g s i n x + c o t x) d x =

Join BYJU'S Learning Program

1∫0etan−1x1+x2 dx is equal to

The correct option is D eπ4−11∫0etan−1x1+x2 dx Let tan−1x=t ⇒11+x2dx=dt =π4∫0et dt=et∣∣π40=eπ4−1

$1 \int 0 \frac{e^{{tan}^{- 1} x}}{1 + x^{2}} d x$ is equal to

The correct option is D $e^{\frac{π}{4}} - 1$
$1 \int 0 \frac{e^{{tan}^{- 1} x}}{1 + x^{2}} d x$
Let ${tan}^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} d x = d t$
$= \frac{π}{4} \int 0 e^{t} d t = e^{t} {∣ ∣}_{0}^{\frac{π}{4}} = e^{\frac{π}{4}} - 1$