Consider the given integral. #10; #10; I=∫t^2t^2+1dt #10; #10; I=∫t^2+1 1t^2+1dt #10; #10; I=∫t^2+1t^2+1dt ∫1t^2+1dt #10; #10; ...

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

Integration by Substitution

∫t21+t2dt =

Question

$\int \frac{t^{2}}{1 + t^{2}} d t =$

Open in App

Solution

Consider the given integral.

$I = \int \frac{t^{2}}{t^{2} + 1} d t$

$I = \int \frac{t^{2} + 1 - 1}{t^{2} + 1} d t$

$I = \int \frac{t^{2} + 1}{t^{2} + 1} d t - \int \frac{1}{t^{2} + 1} d t$

$I = \int 1 d t - \int \frac{1}{t^{2} + 1} d t$

$I = t - {tan}^{- 1} t + C$

Hence, this is the answer.

Suggest Corrections

Similar questions

\int_{1 / e}^{tan x} \frac{t}{1 + t^{2}} d t + \int_{1 / e}^{cot x} \frac{d t}{t (1 + t^{2})}

lim x \to \infty \int_{0}^{x / 2} \frac{t^{2}}{x^{2} (1 + t^{2})} d t

I_{1} = \int_{1 / e}^{tan x} \frac{t}{1 + t^{2}} d t

I_{2} i s \int_{1 / e}^{cot x} \frac{d t}{t (1 + t^{2})}

I_{1} + I_{2}

\int \frac{log (t + \sqrt{1 + t^{2}})}{\sqrt{1 + t^{2}}} d t = \frac{1}{2} (g (t))^{2} + C

g (2)

The points of extremum of the function $F (x) = \int_{1}^{x} e^{- t^{2} / 2} (1 - t^{2})$ dt are

Join BYJU'S Learning Program