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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Finding an Indefinite Integral In Exercises 17-38, find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.)

t ln ( t + 1 ) d t

To determine

To calculate: The value of indefinite integral tln(t+1)dt.

Explanation

Given Information:

The provided indefinite integral is tln(t+1)dt.

Formula used:

The integration by parts udv=uvvdu.

Where, u and v are function of t.

1t+adt=ln|t+a|+C,tndt=tn+1n+1,n1

Calculation:

Consider the indefinite integral tln(t+1)dt

Here,

dv=tdt and u=ln(t+1)

First find v,

dv=tdtdv=tdt

On further solving,

v=t22 …...…... (1)

Find du:

u=ln(t+1)

Differentiate both side with respect t;

dudx=dln(t+1)dtdudt=1t+1

And,

du=1t+1dt …...…... (2)

Apply integration by parts formula and substitute equation (1) and (2) in udv=uvvdu,

tln(t+1)dt=t22ln(t+1)12t2t+1dt=t22ln(t+1)12(t1+1t+1)dt

Apply the formula 1t+adt=ln|t+a|+C,tndt=tn+1n+1,n1,

Therefore,

tln(t+1)dt=t22ln(t+1)12(t1+1t+1)dt=t22ln(t+1)12[t22t+ln|t+1|]+C=t2</

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