Evaluate the integral. ∫x+3x−1dx

Question

Evaluate the integral.  ∫x+3x−1dx

Lible1953 · Accepted Answer

Step 1  Given: I=∫x+3x−1dx  For evaluating given integral, first we simplify given expression then integrate it  Step 2  So,  I=∫x+3x−1dx  =∫(x−1+4)(x−1)dx  =∫(x−1x−1+4x−1)dx  =∫(1+4x−1)dx  =∫dx+4∫dxx−1  (∵∫dx=x+c,∫dxx−a=ln|x−a|+c)  =x+4ln|x−1|+c  Hence, given integral is equal to (x+4ln|x−1|+c).

Charles Clute · Accepted Answer

Step 1
Use Integration by Substitution.
Let u=x-1, du=dx, then x+3dx=u+1+3 du
Step 2
Using u and du above, rewrite

\int \frac{x + 3}{x - 1} d x

.

\int (u + 1 + 3) \times \frac{1}{u} d u

Step 3
Use Sum Rule:

\int f (x) + g (x) d x = \int f (x) d x + \int g (x) d x

.

\int 1 d u + \int \frac{4}{u} d u

Step 4
Use this rule:

\int a d x = a x + C

.

u + \int \frac{4}{u} d u

Step 5
Use Constant Factor Rule:

\int c f (x) d x = c \int f (x) d x

.

u + 4 \int \frac{1}{u} d u

Step 6
The derivative of

\ln x i s \frac{1}{x}

.

u + 4 \ln u

Step 7
Substitute u=x−1 back into the original integral.

x - 1 + 4 \ln (x - 1)

Step 8
Add constant.

x - 1 + 4 \ln (x - 1) + C

Step 9
Merge numbers into the constant.

x + 4 \ln (x - 1) + C

Evaluate the integral. \int \frac{x+3}{x-1}dx

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