Question: "Evaluate the following integral. ∫tt2+1dt"

facas9

Answered question

2021-09-22

Evaluate the following integral.

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

Tasneem Almond

Skilled2021-09-23Added 91 answers

Step 1
Consider the integrals:

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

...(1)
Let,

u = t^{2} + 1

Differentiating w.r.t.t
du=2tdt+0
du=2tdt

t \cdot d t = \frac{1}{2} d u

Step 2
Substitute

u = t^{2} + 1

and

t \cdot d t = \frac{1}{2} d u

in equation (1)
then,

\int t \sqrt{t^{2} + 1} d t = \int \sqrt{u} \cdot \frac{1}{2} d u

= \frac{1}{2} \int u^{\frac{1}{2}} d u

= \frac{1}{2} [\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C

= \frac{1}{2} [\frac{2 u^{\frac{3}{2}}}{3}] + C

= \frac{1}{3} u^{\frac{3}{2}} + C

Back substitute

u = t^{2} + 1

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

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