Consider #10;the following Equation. #10; #10; #160; I=∫_^t^2t^3+1dt #10; #10; #10; #160; x=t^3+1 #10; #10; #160; dxdt=2t ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫t2t3+1 dt =

Question

$\int \frac{t^{2}}{t^{3} + 1} d t =$

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Solution

Consider the following Equation.

$I = \int \frac{t^{2}}{t^{3} + 1} d t$

$x = t^{3} + 1$

$\frac{d x}{d t} = 2 t^{2}$

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

$= \frac{1}{2} ln x + C$

$= \frac{1}{2} ln (t^{3} + 1) + C$

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Similar questions

Q. Solve:-

\int t \sqrt{\frac{t^{2} + 1}{t^{2} - 1}} d t

\int \frac{2 t}{t^{2} - 1} d t

\int \frac{(t^{2} + 1)^{2}}{t^{6} + 1} d t

Chose the correct option where

T_{1}, T_{2} a n d T_{3}

are three different temperatures in case of a liquid.

Let $θ \in (0, \frac{π}{4}) a n d t_{1} = (t a n θ)^{t a n θ}, t_{2} (t a n θ)^{c o t} t_{3} = (c o t θ)^{t a n θ} a n d t_{4} (c o t θ)^{t a n θ}$ , then

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∫t2t3+1dt=

Consider the following Equation. I=∫t2t3+1dt x=t3+1 dxdt=2t2 =∫t2x∗12t2dx =12lnx+C =12ln(t3+1)+C

$\int \frac{t^{2}}{t^{3} + 1} d t =$

Consider the following Equation.

$I = \int \frac{t^{2}}{t^{3} + 1} d t$

$x = t^{3} + 1$

$\frac{d x}{d t} = 2 t^{2}$

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

$= \frac{1}{2} ln x + C$

$= \frac{1}{2} ln (t^{3} + 1) + C$