"Find the indefinite integral. ∫[z3+1(3−z)8]dz" was answered by our community

Yulia

Answered question

2021-11-09

Find the indefinite integral.

\int [z^{3} + \frac{1}{{(3 - z)}^{8}}] d z

Nichole Watt

Skilled2021-11-10Added 100 answers

Step 1
We know that,

\int x^{n} d x = \frac{x^{(n + 1)}}{n + 1} + C

Step 2
We have,

I = \int [z^{3} + \frac{1}{{(3 - z)}^{8}}] d z

= \int z^{3} d z + \int \frac{1}{{(3 - z)}^{8}} d z

= \frac{z^{4}}{4} + \int {(3 - z)}^{- 8} d z

Let us consider,
3-z=t
-dz=dt
dz=-dt

I = \frac{z^{4}}{4} + \int {(3 - z)}^{- 8} d z

= \frac{z^{4}}{4} + \int t^{- 8} (- d t)

= \frac{z^{4}}{4} - \int t^{- 8} d t

= \frac{z^{4}}{4} - \frac{t^{- 7}}{(- 7)}

= \frac{z^{4}}{4} + \frac{1}{7 t^{7}} + C

Now Substituting value of t=3−z, we get our result as

I = \frac{z^{4}}{4} + \frac{1}{7 t^{7}} + C

I = \frac{z^{4}}{4} + \frac{1}{7 {(3 - z)}^{7}} + C

Hence, value of integral

\int [z^{3} + \frac{1}{{(3 - z)}^{8}}] d z i s \frac{z^{4}}{4} + \frac{1}{z {(3 - z)}^{7}} + C

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