Question: "Evaluate the indefinite integral. ∫2x2+x(4x3+3x2)2dx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

aramutselv

Answered question

2021-12-27

Evaluate the indefinite integral.

\int \frac{2 x^{2} + x}{{(4 x^{3} + 3 x^{2})}^{2}} d x

Answer & Explanation

recoronarrv

Beginner2021-12-28Added 20 answers

Step 1
The given integral is,
$I = \int \frac{2 x^{2} + x}{{(4 x^{3} + 3 x^{2})}^{2}} d x$
Using the substitution method, let $4 x^{3} + 3 x^{2} = t$
on differentiating with respect to x on both side of the equation, we get
$\frac{d}{d x} (4 x^{3} + 3 x^{2}) = \frac{d t}{d x}$
$4 (3 x^{2}) + 3 (2 x) = \frac{d t}{d x}$
$12 x^{2} + 6 x = \frac{d t}{d x}$
$6 (2 x^{2} + x) = \frac{d t}{d x}$
$2 x^{2} + x = \frac{1}{6} \frac{d t}{d x}$
Therefore, the given integral becomes,
$I = \int \frac{1}{6 t^{2}} d t$
Step 2
Using the power rule of integration, on solving the given integral, we get
$I = \int \frac{1}{6 t^{2}} d t$
$= \frac{1}{6} \int t^{- 2} d t$
$= \frac{1}{6} (\frac{t^{- 2 + 1}}{- 2 + 1}) + C$
$= - \frac{1}{6 t} + C$
On putting back the value of t, we get
$I = - \frac{1}{6 (4 x^{3} + 3 x^{2})} + C$
Therefore, the expression of the given integral is $- \frac{1}{6 (4 x^{3} + 3 x^{2})} + C$

Chanell Sanborn

Beginner2021-12-29Added 41 answers

\int \frac{2 x^{2} + x}{{(4 x^{3} + 3 x^{2})}^{2}} d x

= \frac{1}{6} \int \frac{1}{u^{2}} d u

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

= - \frac{1}{u}

\frac{1}{6} \int \frac{1}{u^{2}} d u

= - \frac{1}{6 u}

= - \frac{1}{6 (4 x^{3} + 3 x^{2})}

= - \frac{1}{6 (4 x^{3} + 3 x^{2})} + C

Lets

Vasquez

Expert2022-01-07Added 669 answers

$\begin{matrix} G i v e n : \\ \int \frac{2 x^{2} + x}{(4 x^{3} + 3 x^{2})^{2}} d x \\ \int \frac{1}{6 t^{2}} d t \\ \frac{1}{6} * \int \frac{1}{t^{2}} d t \\ \frac{1}{6} * (- \frac{1}{t}) \\ \frac{1}{6} * (- \frac{1}{4 x^{3} + 3 x^{2}}) \\ S i m p l i f y \\ - \frac{1}{24 x^{3} + 18 x^{2}} \\ A d d C \\ A n s w e r : \\ - \frac{1}{24 x^{3} + 18 x^{2}} + C \end{matrix}$

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